- Rules. It defines input and output
main() - it is a function from where compiler starts reading the code. it is the Entry point of the code.
int main() - main() has to return something as output and it always returns int (integer type)
invalid in C++ (only int main is allowed) int main() { } ✔ float main() { } ❌ double main() { } ❌ char main() { } ❌ bool main() { } ❌
void main() - a function which can do something but not need to return
using namesapce std; - it is a shortcut as we don't want to write std:: always
{} - beginning and end of main fuction or the main code which will get compiled
return 0; - tells the compiler / OS I have done writing the code and it is the end of it
endl; - executes the 1st code and takes 2nd code in new line. NEW LINE
strings are written in double quotes " "
int main() {
int num = 10; num is a variable and we can write anything like int BC = 10; and it will retuen 10 as integer
}
int main() { int anyvariablename = 10; cout << anyvariablename; // output will be 10 return 0; }
Integer-
Integer range-> 10^-9 to 10^9
int main() {
int numInt = 10; // no double quotes in variable name
cout << "Integer is: " << numInt << endl;
return 0;
}
Output- Integer is: 10
Long-
Long range-> 10^-12 to 10^12
#include using namespace std;
int main() { int numInt = 10; cout << numInt << endl;
long numLong = 1000000000;
cout << numLong;
return 0;
}
Output- 10 1000000000
cout << INT_MAX << endl; cout << LONG_MAX << endl; cout << LLONG_MAX << endl; (after including )
#include <iostream>
#include <climits>
using namespace std;
int main() {
cout << INT_MAX << endl;
cout << LONG_MAX << endl;
cout << LLONG_MAX << endl;
return 0;
}
Output-> 2147483647 2147483647 9223372036854775807
- its better to use '\n' compared to "\n" although the difference is tiny
- when we cout << "Hello"; the text is not printed immediately on terminal, instead it goes on temporary memory area called Buffer
- Flushing means take everything currently sitting in the buffer and send it to the output device right now. cout << "Hello"; cout.flush();
- endl is equivalent to print a newline & flushing the buffer
- although flush or endl is better but it is not free, mostly '\n' is used
- buffer is like a truck carrying packages
LONG LONG-
range of long long -> 10^-18 to 10^18
int main() { int numInt = 1000000000; cout << numInt <<endl;
long long numLong = 1000000000000000000;
cout << numLong;
return 0;
}
- Number data Types-> int , long , long long
- Decimal number data types-> float , double
- single alphabet/letter/symbol-> char
- multiple letters-> string (string is a class under std & not data tyoe, Eg- "Sarbesh can do it!")
- true/false-> bool
float & double-
float store decimal points like 10.7 Range of float -> we can store upto 7 decimal places Range of double float -> upto 15 decimal places
int main() { float numFloat = 8.7; cout << numFloat; return 0; }
Output- 8.7
char-
char stores a single letter or symbol like a single letter of "a" or "$" or "Z"
int main() { char ch = 'a'; cout << ch; return 0; }
Output- a
int main() { char ch = 'a'; cout << "Character :" << ch << endl; return 0; }
Output- Character : a
string-
string is a class under std and its not a data type and it stores collection of characters
#include using namespace namespace std;
int main() { string str = "Sarbesh can do it!"; cout << str;
return 0;
}
Output- Sarbesh can do it!
bool
boolean stores only true & false bool
Combining-
int main() { int age = 25; float height = 5.9; char grade = 'A'; bool isStudent = true; string name = "Rahul"; return 0; }
cin reads only one word . stops at space, newline, tab
getline() reads the entire line including spaces
char reads only first letter
- Input/Output-
output -- ask user to put value int main() input -- store it int num1, num2; // one data type then multiple variables output operations -- show it as result
tuf eg code-
int main() {
int age;
// Output: Ask user a question
cout << "Enter your age: ";
// Input: Read what user types and store in 'age'
cin >> age;
// Output: Show the result
cout << "Your age is: " << age;
return 0;
}
Dynamic input code-
int main() {
int num1, num2, num3, num4; // we are taking 4 variables here
// cout << "Enter 4 numbers: " << "\n";
cin >> num1 >> num2 >> num3 >> num4;
cout << num1 + num2 << endl;
cout << num3 + num4;
return 0;
}
input-output4- (taking 2 variables i.e bdate & num)
int main() {
int bdate, num;
cin >> bdate;
cin >> num; // alternatively we can write cin >> bdate >> num;
cout << "The date of birth is :" << bdate << endl;
cout << "The number is :" << num;
return 0;
}
input-output5- (again taking 2 variables but asking user output first)
int main() {
int bdate; // Alter int bdate, num;
int num;
cout << "Enter your DoB : ";
cin >> bdate;
cout << "Enter a number : ";
cin >> num;
cout << "The DoB is : " << bdate << endl;
cout << "The number is : " << num;
return 0;
}
input-output6- (Taking 2 data types- int & float)
int main() {
int bdate; float num;
cout << "Enter your DoB : "; cin >> bdate;
cout << "Enter a number : "; cin >> num;
cout << "The DoB is : " << bdate << endl << "The number is : " << num;
return 0;
}
/* Input- Enter your DoB : 22 Enter a number : 7.85
Output- The DoB is : 22 The number is : 7.85
// to show we dont hv to write so many lines */
- if/else-
if is a conditional statement if()
and we close braces {} when we start if and end it
if(...) it is true or the conditions are met then everything written in the curly braces will be performed.
a condition is expressed in true or false 10 > 5 → true 10 < 5 → false a == b → true/false
if(condition)
Struct of if- if (....) { }
if(conditions) { if conditions are true then do logic/operations/print }
else- if the previous condition was false do this instead else don't have any condition or logic
think of this as- if (rain) take umbrella else don't take umbrella
we use else if when there are multiple conditions
Marks >= 90 → Grade A Marks >= 75 → Grade B Marks >= 50 → Grade C Otherwise → Fail
The compiler checks from top to bottom. The first true condition wins. After finding a true condition, it skips the rest.
if(x > 90) if(x > 90) if(x > 75) else if(x > 75) --> They are not same as if's all conditions are checked and with else if checking stops when the 1st true condition arrives if(x > 50) else if(x > 50)
Structure- if(conditionals) { operation if condition is true and checks all the next if's condition }
else { just operation when if is false }
else if (conditions) { operation and stops at 1st true condition }
if & else-tuff code-
Problem statement- if the user has >1000 then cout buy a pizza or else if it is less then cout I will buy a burger
int main() {
int money = 500;
cout << "how much balance you have :";
cin >> money;
if (money >= 1000) {
cout << "I will buy a pizza";
}
else {
cout << "I will buy a burger";
}
return 0;
}
if & else code-
Problem Statement- Given an age, print adult >= 18, or print "Teen"
int main() {
int age;
cout << "Enter your age: ";
cin >> age;
if (age >= 18) {
cout << "Adult"; // if we don't use else, then the program will print "Adult" for ages 18 and above, but it will not print anything for ages below 18
}
else {
cout << "Teen";
}
return 0;
}
if-else-multiple-
Problem Statement- // Given an age, // if the age >= 18, print "Adult" // if the age < 18 and >= 10, print "Teen" // if the age < 10, print "Child"
int main() {
int age;
cout << "enter your age: ";
cin >> age;
if (age >= 18) {
cout << "Adult";
}
else if (age < 18 && age >= 10) { // in "and" "&&" operator both the condition should be true to fullfill the overall condition
cout << "Teen"; // or else if we want one condition can be true we will use OR " || " operator
}
else if (age < 10) {
cout << "Child";
}
return 0;
}
if-else-multiple2-
Problem statement- Given the marks of a student, tell us the grade he is getting following the below rules Grade A- (>=90) grade B- (>=70 and <90) Grade C- (>=50 and <70) Grade D- (>=35 and < 50) Fail- (< 35)
int marks;
cout << "Enter marks :";
cin >> marks;
if (marks >= 90) {
cout << "Grade A";
}
else if (marks >= 70 && marks < 90) {
cout << "Grade B";
}
else if (marks >= 50 && marks < 70) {
cout << "Grade C";
}
else if (marks >= 35 && marks < 50) {
cout << "Grade D";
}
else {
cout << "Fail";
}
return 0;
switch case-
it is another way of writing if statement but what can be possible if-else conditions
imagine we have exact values like ( 1 = Monday or 90 = Grade A ) and not marks > 90 then we need if-else it works best when exact value matches but we need breaks or else program keeps flowing downwards until switch ends and that's called Fall-through break; means exit the switch immediately
default means like else in if-else chain if no case mathces- default: runs
switch() {
}
switch(variable name) { // switch(day) {}
case 1:
break;
case 2: break; }
switch-case code-
int main() {
int day;
cout << "Enter the day: ";
cin >> day;
switch(day) {
case 1:
cout << "Monday";
break;
case 2:
cout << "Tuesday";
break;
case 3:
cout << "Wednesday";
break;
case 4:
cout << "Thursday";
break;
case 5:
cout << "Friday";
break;
case 6:
cout << "weekend";
break;
case 7:
cout << "weekend";
break;
default:
cout<< "Invalid";
}
return 0;
}
What is Nested if-else? A nested if-else means: 👉 an if statement inside another if statement. So first outer condition checks, then inner condition checks. Example: if (condition1) {
if (condition2) {
// code
}
}
Important Learning Outer if - Narrows down possibilities. Inner if - Does more detailed checking. That is exactly how nested conditions work.
Problem statement- You are given three integers a, b and c print which of these integers are largest if two or more integers are equal and are the largest, print any of them the program should indicate that as well , user will input the 3 integers at runtime
int main() {
int a, b, c;
cout << "Enter your 3 numbers: ";
cin >> a >> b >> c;
if (a >= b) {
if(a >= c){
if (a == b || a == c) {
cout << "largest number is: " << a << " & There's a tie" << endl;
}
else {
cout << "largest number is: " << a << endl;
}
}
else {
cout << "largest number is: " << c << endl;
}
}
else if ( b >= c) {
if ( b == c) {
cout << "largest number is: " << b << " & There's a tie" << endl;
}
else {
cout << "largest number is: " << b << endl;
}
}
else {
cout << "largest number is: " << c << endl;
}
return 0;
}
Problem statement- You are given three integers a, b and c print which of these integers are largest if two or more integers are equal and are the largest, print any of them the program should indicate that as well the 3 integers are 36 56 78
int main() {
int a = 36;
int b = 56;
int c =78;
if (a >= b) {
if (a >= c) {
cout << "Largest is a";
}
else {
cout << "Largest is c";
}
}
else if (b >= c) {
cout << "Largest is b";
}
else {
cout << "Largest is c";
}
return 0;
}
- So first outer condition checks, then inner condition checks.
Visual Structure-
if (a >= b)
if (a >= c)
a largest
else
c largest
else
if (b >= c)
b largest
else
c largest
This is the core nested if-else pattern.
Cleaner code with no ties-
if (a >= b) {
if (a >= c)
cout << a;
else
cout << c;
} else {
if (b >= c)
cout << b;
else
cout << c;
}
'=' and '==' both are different
= is an assignment operator, int a = 10; meaning store value 10 in variable a
== is an equality operator, it checks whether two values are equal or not, if (a == 10) means check if a = 10 or not
- Loops-
- when certain no. of things are repeated use LOOPS
- first one is initializer, 2nd one is condition or break statement, third one is increment/decrement
Problem Statement- take input of 10 numbers and print it OR Print all the numbers from 1 to 10
here we can use Loops instead of writing 10 nos
if we are not writing loops then- int main() { int num1, num2, num3, num4...num10; cout << "enter no"; cin >> num1 >> num2 >> ...num10; }
int main() {
for (int i = 1; i <= 10; i++) {
cout << i << endl;
}
return 0;
}
Output- 1 2 3 4 .. 10
for (int i = 1; i = i + 1) - no condition, always checks in. This is INFINITE LOOP , prog dosen't know when to stop and keep revolving around for & will never reach last condition that is return 0;
for (int i = 1; i<=2;) - Again Infinite loop as I failed to give increment then aslo it will not stop because i will always be 1 and 2nd condition is always be true
for (int i = 1;true;i++;) - Again Infinite loop
that's why you need to give valid BREAK'S statement which gonna be false at some moment
for(initialization; condition; update) for(int i = 1; i <= 10; i++)
// Problem statement- print hello 5 times using loops
int main() {
int i;
for (int i = 1; i <= 5; i++) {
cout << "Hello" << endl;
}
return 0;
}
Problem Statement- Print odd numbers from 1 to 20
int main() {
for (int i = 1; i <= 20; i = i+2) {
cout << i << endl;
}
return 0;
}
Problem statement- Print even numbers from 1 to 20
int main() {
for (int i = 2; i <= 20; i = i+2) {
cout << i << endl;
}
return 0;
Problem stat- print 1 to 10 numbers but in reverse order
int main() {
for( int i = 10; i >= 1; i--) {
cout << i << endl;
}
return 0;
}
problem stat- print all numbers from 10 to 1 in reverse order but print just even numbers
int main() {
for (int i = 10; i >= 1; i = i - 2) {
cout << i << endl;
}
return 0;
}
Problem stat- Print all the multiples of 5 till 100
int main() {
for (int i = 5; i <= 100; i = i + 5) {
cout << i << endl;
}
return 0;
}
Problem Statement- Take 5 inputs and print their sum Problem summary - This program uses a loop to take 5 numbers one by one, keeps adding them to sum, and after all inputs are taken, prints the final total
- It's an accumulation pattern
int main() {
int num;
int sum = 0;
cout << "Enter 5 nos: ";
for(int i = 1; i <= 5; i++) {
cin >> num;
sum = sum + num;
}
cout << "Sum = " << sum;
return 0;
}
Input: 2 4 6 8 10 output: Sum = 30
int sum = 0; - we create a variable to store the local initially sum = 0 Loop runs 5 times - for(int i = 1; i <= 5; i++) take input - cin >> num; add number to the sum - sum = sum + num;
Example: sum = 0 user enters 5 new sum = 0 + 5 = 5
Then: user enters 3 new sum = 5 + 3 = 8 and so on.
instead of sum = sum + num , u can write- sum += num;
-
int num; -> Its a temporary storage box. Every time the user enters a number, it goes into num Example: User enters 2 → num = 2 User enters 4 → num = 4 User enters 6 → num = 6 -num keeps getting overwritten with the latest input.
-
sum stores the total. We start with zero.
-
for(int i = 1; i <= 5; i++)
-
this means take 5 inputs one by one just like a loop instead of writing manually everytime: cin >> num; cin >> num; cin >> num; cin >> num; cin >> num;
we use a loop to repeat the same task 5 times.
- here i is just used for counting.
- We are not printing i, and we don't really care about its value. We only use it to make the loop run 5 times.
Q. Why is it called "Accumulation Pattern"? A. Because a variable keeps accumulating (collecting) values over time.
int num; int sum = 0;
num → stores each number entered by the user. sum → stores the running total. We initialize sum to 0 because we haven't added anything yet.
Think of sum as a bucket: sum = 0 Every new number gets added into the bucket.
Loop for(int i = 1; i <= 5; i++)
This loop runs 5 times.
| Iteration | i |
|---|---|
| 1st | 1 |
| 2nd | 2 |
| 3rd | 3 |
| 4th | 4 |
| 5th | 5 |
Taking Input cin >> num; // Stores one number in num.
Example: User enters 2 num = 2
Accumulating the Sum sum = sum + num;
This means: Take the old value of sum and add the new number to it.
Example if user enters: 2 4 6 8 10
| Input | Calculation | sum |
|---|---|---|
| 2 | 0 + 2 | 2 |
| 4 | 2 + 4 | 6 |
| 6 | 6 + 6 | 12 |
| 8 | 12 + 8 | 20 |
| 10 | 20 + 10 | 30 |
Final: sum = 30
Output cout << "Sum = " << sum;
Prints: Sum = 30
Dry Run-
Suppose input is: 1 2 3 4 5
Start: sum = 0
Iteration 1: num = 1 sum = 0 + 1 = 1
Iteration 2: num = 2 sum = 1 + 2 = 3
Iteration 3: num = 3 sum = 3 + 3 = 6
Iteration 4: num = 4 sum = 6 + 4 = 10
Iteration 5: num = 5 sum = 10 + 5 = 15
Output: Sum = 15
we wrote " cout << i << endl; " in every other loop programs why not here?
in earlier programs- for(int i = 1; i <= 5; i++) { cout << i << endl; }
Here the goal is Print every value of i. So during each iteration, you immediately display i Output: 1 2 3 4 5
But in this program: for(int i = 1; i <= 5; i++) { cin >> num; sum = sum + num; }
The goal is different: Add all 5 numbers together and print only the final answer.
if we write " cout << sum << endl; " inside the loop we will get Output: 2 6 12 20 30
because sum changes after every input.
But our probelm asked for final number after all the numbers have been added Thats why cout << "Sum = " << sum; is placed outside the loop
this is what i was talking about being inside the for loop:- for(int i = 1; i <= 5; i++) {
cin >> num; sum = sum + num; cout << "Sum = " << sum << endl; }
Enter 5 nos: 2 4 6 8 10 Sum = 2 Sum = 6 Sum = 12 Sum = 20 Sum = 30
Notice how the meaning changes based on where cout is placed:
// Inside loop for(...) { ... cout << sum; }
➡️ Print after every iteration.
for(...) { ... }
cout << sum;
➡️ Print only the final result after the loop finishes.
This is one of the most important ideas in loops: statements inside the braces {} run every iteration; statements outside run only once.
A picture of when i is inside and outside the loop and i is not only used for counting
The first 1 2 3 4 5 come from inside the loop. Final 6 comes from outside the loop
i is usually used as a loop counter, but it can also be used for other purposes if needed.
Remember loop does not stop at 5. It stops when the condition becomes false.
While Loop-
- Another syntatcical way of writing for loop
- never forget to include update or increment/decrement in while loop (i++) orelse it will be an infinite loop
int i = 1;
while(i <= 5) { cout << i << endl; i++; }
for(int i = 1; i <= 5; i++) { cout << i << endl; }
Problem stat- Print all the multiples of 5 till 100
int main() {
int i = 5; // Initializer
while( i <= 100) { // Condition
cout << i << endl; // Operations
i = i + 5; // Increase
}
return 0;
}
int main() { for(int i = 5; i <= 100; i = i+5) { cout << i << endl; }
Whenever you write a while loop, always identify these three parts:
Initialization → Where does it start? Condition → When should it continue? Update → How does it move toward stopping?
If any one of these is wrong, the loop either gives wrong output or becomes an infinite loop.
A image of i inside the loop and outside the loop where 105 is printed after 100 as i called i outside the loop also and then the condition got failed
- Arrays-
An Array is an multiple values of the same type stored in contiguous memory locations. Arrays are containers where we can keep similar data types.
For an array of size 5- Index: 0 1 2 3 4 Value: 10 20 30 40 50
First index = 0 Last index = n - 1
For size 5 , First index = 0, Last index = 4 For size 10, First index = 0, Last index = 9
Accessing elements- arr[0] arr[2] arr[4]
arr[2] is the index position
arr[5] = {10, 20, 30, 40, 50}; // example
Think of an array as a row of boxes:
Index: 0 1 2 3 4
+----+----+----+----+----+
Array: | 10 | 20 | 30 | 40 | 50 |
+----+----+----+----+----+
Now, arr[0] → element at index 0 → 10 arr[2] → element at index 2 → 30 arr[4] → element at index 4 → 50 arr[5] -> dosen't exists
int arr[5];
- it means size of 5 -> the array that can store 5 elements
- Valid indices = 0 to 4 So: ✅ arr[0] ✅ arr[1] ✅ arr[2] ✅ arr[3] ✅ arr[4] ❌ arr[5] (invalid, because index 5 doesn't exist)
arr[2] means Give me the element stored at index 2.
for(i = 0; i < n; i++) Think that, I am visiting every element exactly once
Suppose size is 5 -> int arr[5] arr[0] arr[1] arr[2] arr[] -> This is Index arr[3] arr[4] This is Valid
arr[5] This is Invalid. This is called Out of bounds access
num[0] -> 0 positions away from start num[1] -> 1 position away from start num[2] -> 2 positions away from start
- [] -> array initilization. int arr[] , int num[] , int sarbesh[] all works.
First index = 0 Last index = size - 1 Number of elements = size
first index = 0 last index = size - 1
data_type array_name[size] = {};
int arr[5] = {..};
There are three parts: int arr[5]; │ │ │ │ │ └── Size │ └────── Array name └────────── Data type
- I can name any array name according to my choice int sarbesh[5]; int marks[100]; int ages[10]; int scores[20];
cout << sarbesh[4]; -> output element of index 4 or the last element
- Array can store any data type also & not only int long population[10]; float temperature[7]; double salary[20]; char grade[5]; bool passed[50]; string names[10]; // string array
data_type array_name [size] │ │ │ What it stores Its name Capacity
if we need different numbers or different variable sunder same data type what's the point of writing int num1, num2, num3... when we write int num; - it picks up a memeory location & whatever value u assigns it get stored in that memory data type variable [how many elements that array have] - int num[5]; - it figures out 1st,2nd,3rd,4th,5th memory address & binds them & keep it contagious - that's 5 contagious memory location
- every memory location can store an integer & we hv 5 memory locations & all of 5 locations can store integer
int main() {
int num = 5;
cout << # // when we type &num it shows where the 5 is stored in memory in this case it is 0x61ff0c
return 0; // Output- 0x61ff0c - this is memory address where 5 is stored
}
data type variable [how many elements that array have] - int num[5]; - it figures out 1st,2nd,3rd,4th,5th memory address & binds them & keep it contagious - that's 5 contagious memory location
- every memory location can store an integer & we hv 5 memory locations & all of 5 locations can store integer
A small terminology tip: People often get confused because of counting:
Array size = 5 Elements = 5 Indices = 0, 1, 2, 3, 4
So: The 5th element is num[4]. ✅ num[5] is actually trying to access the 6th element, which does not exist. ❌
& -> shows the memory location & -> The & operator is called the address-of operator.
let's say datatype of int num = 5; cout << # -> it shows the memory location of variable num stored as integer data type
how int a = 10; and int b[] = {10}; are different
- int a = 10 means create one box & store 10
- Create an array with one box, and store 10 in the first (and only) element.
- Variable -> One Box , no index needed
- Array -> Many boxes , index needed
Output diff: cout << a; // 10 cout << b[0]; // 10
int marks[] = {90, 85, 78}; equals to int marks[3] = {90, 85, 78};
- Remember array size = elements
- [] = {}
- index is -1 always
zero filling -> 
mistake to avoid -> 
trick to remember -> 
For an array of size n: for (int i = 0; i < n; i++)
Problem statement- Create an array of 5 integers. Ask the user to enter numbers. Store each number in the correct index. Immediately print the number that was entered
int main() {
int num[5];
cout << "Enter 5 numbers: ";
for (int i = 0; i < 5; i++) {
cin >> num[i];
cout << num[i] << endl;
}
return 0;
}
Input-> 10 20 30 40 50
Output- 10 20 30 40 50
Problem Statement- Take 5 numbers from us and print them and store it in an Array
int main() {
int num[5];
cout << "Enter 5 numbers: ";
for(int i = 0; i <= 4; i++) {
cin >> num[i];
cout << num[i] << "\n";
}
return 0;
}
Because arrays start from index 0 So 0 1 2 3 4 Total = 5 elements
Important Concept-
int num[5]; // Creates 5 boxes
| Index | Value |
|---|---|
| 0 | ? |
| 1 | ? |
| 2 | ? |
| 3 | ? |
| 4 | ? |
When user enter values: it get stored inside-- cin >> num[i]; // user input gets stored in those boxes
if we want our output to come horizontally we have to use " " cout << num << " ";
for vertical output we use endl or "\n"
- Strings-
we use string instead of char because char allows single letter or single symbols and string allows multiple letters char -> Stores one character string -> stores a sequece of characters
- Strings are basically Arrays of characters + extra functionality
Array: [10,20,30,40]
String: ['H','e','l','l','o']
Indexing Traversal Length Access by position
cin stops reading after space
Suppose, user enters Sarbesh Mallick as input
Using Cin ->
string name;
cin >> name;
Output stored in name: Sarbesh
not entire name- Sarbesh Mallick because Because cin stops reading when it encounters:
- space
- tab
- newline
So, Sarbesh Mallick ^ stop here , Mallick remains in the input buffer
string name;
getline(cin, name);
Output stored in name: Sarbesh Mallick (full name)
Quick Rule- cin -> reads one word getline() -> reads the entire line
string name = "Sarbesh";
Character: S a r b e s h Index: 0 1 2 3 4 5 6
name[0] -> S (1st character) name[6] -> h (last character)
What if -> name[7] ?
for Sarbesh
- Length = 7
- Last Index = 6
So, name[7] is out of bounds and should be considered invalid access.
This is the exact same rule you learned for arrays:
Length = n Valid indices = 0 to n-1
For "Sarbesh":
Length = 7 Valid indices = 0 to 6
It means Visit every element one by one.
For an array of [10, 20, 30, 40, 50] Traversal means:- Visit 10 Visit 20 Visit 30 Visit 40 Visit 50
For a string: Sarbesh traversal means:- Visit 'S' Visit 'a' Visit 'r' Visit 'b' Visit 'e' Visit 's' Visit 'h'
when we write for(int i = 0; i < name.length(); i++) Think of i as a finger moving across the string. i=0 for S then i=1 for a then i=2 for r and so on..... The loop variable is literally walking through the string.
Why do we traverse?
- Because we usually want to do something with each character.
For Array:- 10 20 30 40 50 Traversal:- arr[0] arr[1] arr[2] arr[3] arr[4]
For String:- Sarbesh Traversal:- name[0] name[1] name[2] name[3] name[4] name[5] name[6]
Array -> elements are integers / numbers String -> elements are characters
vairable name.lenth()
Eg:- cout << fullName.length();
Length counts every character including spaces For, Sarbesh Mallick:- S a r b e s h -> 7 (space) -> 1 M a l l i c k -> 7 Total:- 7 + 1 + 7 = 15
So, fullName.length() returns 15
fullName.length() is same as fullName.size() and for string it does the same thing.
For, Sarbesh Mallick Length -> 15 Last Index -> 14 (Last Index = Length - 1)
.length()
variablename.length();
variablename.size();
.lenght() = .size() for strings but, .size() is a general function for many c++ containers but .length() exclusively for characters
- vector has .size()
- array has .size()
- string has both .size() and .length()
[] → "Which character?" to print on every iteration .length() → "How many characters?"
- 1st getline , 2nd getline
- 1st getline , 2nd cin
- 1st cin , 2nd getline ❌ cin.ignore() required
Correct way of writing 3rd condition- string first, last; cin >> first; cin.ignore(); getline(cin, last);
| Previous input | Next input | Need cin.ignore()? |
|---|---|---|
getline() |
getline() |
❌ No |
cin >> |
getline() |
✅ Yes |
getline() |
cin >> |
❌ No |
cin >> |
cin >> |
❌ No |
| Previous input | Next input | Need cin.ignore()? |
|---|---|---|
getline() |
getline() |
❌ No |
cin >> |
getline() |
✅ Yes |
getline() |
cin >> |
❌ No |
cin >> |
cin >> |
❌ No |
Problem Stat- Store 2 strings and print it and show its length and also find the index of 3rd element
int main() {
string firstname, lastname, fullname;
cout << "enter your firstname: ";
getline (cin, firstname);
cout << "enter your lastname: ";
getline (cin, lastname);
fullname = firstname + " " + lastname;
cout << "your fullname is: " << fullname << endl;
cout << "length of fullname is: " << fullname.length() << endl;
cout << "length of firstname is: " << firstname.length() << endl;
cout << "index of 3rd element: " << fullname[3];
return 0;
}
Alternative-> if we want to use fullname.length() multiple times
cout << "length of fullname is: " << fullname.length() << endl; you could store it if you were going to use it multiple times:
int length = fullname.length(); cout << "Length = " << length;
Not necessary here, but useful in bigger programs.
// Probelem Statement- Store 2 strings and print it and show its length and also find the index of 3rd element
#include <iostream>
using namespace std;
int main() {
string firstname = "Sarbesh";
string lastname = "Mallick";
string fullname = firstname + " " + lastname;
cout << "My fullname is: " << fullname << endl;
cout << fullname.length() << endl;
cout << fullname[3];
return 0;
}
Output- My fullname is: Sarbesh Mallick 15 b
- Image->
For example: string city = "Bangalore";
Here: city is the variable name. "Bangalore" is the string value stored inside the variable.
Think of it like a labeled box: Variable Name: city Value: Bangalore
but numbers like integers dosen't need " "
int age = 22;
string name = "Sarbesh";
char grade = 'A';
Notice:
age -> variable name
name -> variable name
grade -> variable name
like in,
string city = "Bangalore";
city is varibale and bangalore is string literal
Single Quotes means one character ' '
Problem stat- let the user type its fullname , store it in string, print it and show its firstname length & fullname length
int main() {
string firstname, lastname, fullname;
cout << "Write your firstname: ";
cin >> firstname; // getline(cin, firstname); - we can write also this to store middlename like Sarbesh Kumar as cin ends after space
cout << "Write your last name: ";
cin >> lastname;
fullname = firstname + " " + lastname;
cout << "My fullname is: " << fullname << endl;
cout << "Lenght of the first name is: " << firstname.length() << endl;
cout << "Lenght of the fullname is: " << fullname.length();
return 0;
}
One thing to be careful about
If there was a previous cin before the getline(), then you may run into the famous: getline() gets skipped problem because cin leaves a newline (\n) in the input buffer.
Example: cin >> age; getline(cin, firstname); // may get skipped
In such cases, you'd use: cin.ignore(); before getline().
for my current input as getline is first there is no problem
int main() {
string str1;
string str2;
cout << "just hit random: ";
getline(cin, str1);
cout << "hit amazing ";
getline(cin, str2);
cout << "Life is Enjoy " << str1 << endl;
cout << "Hit is Enjoy " << str2;
return 0;
}
getline(cin, ?); - // ? is the varibale
#include cin.ignore(numeric_limits::max(), '\n');
- This removes everything up to the next newline, making it more robust if there are extra characters left in the input buffer.
Problem Statement: Given a string, print each character of the string using a for loop. Problem Statement: Traverse a string character by character and print each character. Problem Statement: Given a string, access and print each character using its index. Problem Statement: Store a string and print all its characters by accessing them through their indices.
int main() {
string str = "sarbesh mallick";
for (int i = 0; i < str.size(); i++) {
cout << str[i];
}
return 0;
}
output- sarbesh mallick
str[i] - Give me the character stored at index i , each iteration and together it becames sarbesh str - means the entire string, not a single character. Every iteration prints sarbesh and loop runs 7 times i.e sarbeshsarbeshsarbeshsarbesh....
str[6] - inside the loop & will print 15 times str[7] - inside the loop, it will be blank becasue space btw sarbesh and mallick it is str outside the loop will simply print sarbesh mallick as it is
- Functions-
Imagine you hv-
- take 2 numbers
- Add them
- print result If you need this logic 10 times, you hv to write it 10 times. Instead, Functions let you Write once, reuse many times
main() - entry point of the code & the first function that the operating system calls when your program starts.
- A function is a block of code that does a specific task. We write it once and call it whenever we need it. It's like a "helper" or a "servant". We teach the servant how to make tea once, and then just order "Make Tea" anytime. Functions keep our code clean and reusable
int main() is the entry point of cpp code void main() - it is a self function where it will do something but will not return anything void main() is a non-standard and should not be used, even though some old compilers accept it.
- Now the function contains the logic, and the loop decides how many times to run it
- Function → What to do Loop → How many times to do it
- we can use loop and functions togerther
#include <iostream>
using namespace std;
void print() {
cout << "I am a print function" << endl;
}
int main() {
return 0; // It will return nothing or print nothing , for that we need to call print() after int main()
}
Problem statement- Write a C++ program to create a user-defined function print() that prints a message, and call it from the main() function.
#include <iostream>
using namespace std;
void print() {
cout << "I am a print function" << endl;
}
int main() {
cout << "Before print function call" << endl;
print(); // Whenver the function is called, its get a priority, it will be executed first & then reamining lines
cout << "After print function call" << endl;
return 0;
}
Output- Before print function call I am a print function After print function call
void print() { cout << "I am a print function" << endl; // This does not excute anything, it only tell compilers that there exists a function named print() } // Its like a tool, tool exists but nobody is using it
program always starts from int main() and not from print()
-
so after int main the compiler reads "Before Print function call" and it prints that // Output-Before Print function call
-
now it reads print();
-
when the cpu sees print(); - Pause main() Jump to print() Execute print() Come back / return Continue main()
-
inside print() cpu jumps here- void print() { cout << "I am a print function" << endl; // Output- I am a print function }
-
Functions ends as there is no explicit return 0; but as it is void it automatically returns when it reaches }
-
CPU goes back to the exact place where print()
-
Continue with main()
cout << "After print function call" << endl; // Output- After print function call -
Final Output- Before print function call I am a print function After print function call
When a function is called, program control jumps to that function. After the function finishes, control returns to the caller.
void means it returns nothing, its job is only to print something or doing something void dosen't return values to int main() , it just does it job and shows it and exits like void add() prints the cout but dosen't hand anything to main()
-
void function er modhye sob logic , statements thkbe, I just need to call it and ota execute hbe
-
but returning function e amke return kichu krte hbe to int main() caller and shey decide krbe ki krte hbe returned value niye, shey chaile cout lrte pare ki add
-
eg: void add() { cout << 6; logic... }
add(); ✔ // void function ke call krlam and it will execute whatever logic i had written inside the void add() {} cout << add(); ❌ // not possible -
eg: int add() { cout << 6; return 5; }
add(); // 6 will get printed, cout << add(); // then 6 & 5 both will get printed cout << add() + 5; // 10 will get printed (5+5)
int add() {
cout << 6 << endl;
return 5;
}
int main() {
cout << "Before print function call" << endl;
cout << add() << endl;
cout << "After print functionn call" << endl;
return 0;
}
cout << 6; - Show 6 on the screen return 5; - Give 5 back to the caller & caller will decide what do with it
-
Cpp starts from main()
-
The add() function is not executed yet. Think of add() as a worker waiting for instructions.
-
Execution reaches "Before print function call" and shows the same as output
-
Now, execution reaches cout << add() << endl;
-
cout dosen't print first, it pauses and jumps to see what's the value of add()
-
think of it like: main() ↓ cout << add(); ↓ Wait... I need to know add() ↓ Go to add()
-
As now we are inside int add() { cout << 6 << endl; // prints 6 return 5; // This does NOT print 5. It returns the value to the caller. }
-
now the program comes back to the cout << add() << endl;
-
the computer replaces add() with 5 so it becomes cout << 5 << endl; which prints 5
int square() { return 25; } square(); ❌ // Nothing gets printed, Because nobody is using the returned value. The function simply returns 25, and then it's discarded. cout << square(); ✔ // Now, 25 gets printed
Void Function->
void printName() { cout << "Sarbesh"; } printName(); // Call: printName();
Output: Sarbesh But what value came? -> Nothing cuz void means no return value
Returning Function->
int getAge() { return 22; } cout << getAge(); // Call
Output: 22
Trick Question->
Suppose-
void printName() { cout << "Sarbesh"; }
Can we do- cout << printName(); ❌
Because, printName() returns nothing and cout needs a value to print
Mental Model- cout ↓ Screen
return ↓ Caller Function
- Different destinations.
The function istelf has no special power. like print() dosent print automatically we have to write cout under that so, we can name te function anything like void add() or void hello() and still print something by writing cout the function name is just a label
The important thing is what does the function does inside its body
-
Never write code after return as the function has already exited
-
in void function everything happens inside and main gets nothing, whatever u want do with void u do it inside the void and just call it when reqd.
-
in returning functions, u do the math or whatever and give it to main and main will further do processing or just print it
-
void -> Includes everything under it
-
returning function -> just do basic thing and return to main caller and the main will handle the rest
-
with void whatever we do and if we print results, it just get printed and lost we cannot usally use it later in different way
- when we write returning function, by default it promises it will return something to caller.
- but when we don't return something, and only call the function that's still okay, it executes logic inside the function but dosen't return anything.
- but if we cout << function-name(); then it will provide a garbage value because int function-name() promised to return an integer value which it failed
- a function task is to deliver the return value to main() caller, and main caller will decide what to do
- a return type function will always return something except void
-
int add() { return 5; // returns an integer } cout << add(); // Output- 5
-
int age() { return 20; // returns a character. }
-
double pi() { return 3.14; // returns a decimal number. }
-
string name() { return "Sarbesh"; // returns a string. }
-
char grade() { return 'A'; // returns a character }
But
- void print() { cout << "Hello"; // returns nothing }
Ans- NO
- The code that calls a function receives its return value.
- It doesn't have to be main().
- for example-
#include <iostream>
using namespace std;
int add() {
return 5;
}
void printNumber() {
cout << add(); // Here, printNumber() is calling add()
}
int main() {
printNumber();
return 0;
}
Output- 5 In this case, printNumber() receives the value 5, not main() directly.
My Current function with void-
void twonosprint() {
int num1, num2;
cin >> num1 >> num2;
cout << num1 + num2;
}
- Suppose user enters 10, 20 and Output will be 30 then
- Now, 30 is printed on the screen
- but can another part of program uses 30? -> NO -> cuz it is printed not retuned
- twonosprint(); - can print 30
- twonosprint() * 10 - Can't do ❌ cuz the function returns nothing
void function ↓ Performs an action
Returning function ↓ Produces a value
void function ↓ Usually performs an action (printing, modifying data, etc.)
int function ↓ Usually computes something and returns an answer This isn't a strict rule, but it's a very useful beginner mental model.
- A printed value is visible to the user. A returned value is usable by the program.
C++ follows the same order of operations as Mathematics (BODMAS/PEMDAS).
- Parentheses
() - Multiplication
*, Division/, Modulus% - Addition
+, Subtraction-
cout << num1 + num2 * 10;Input:
num1 = 1
num2 = 2
Evaluation:
1 + 2 * 10
= 1 + 20
= 21
Output: 21
Multiplication is performed before addition.
cout << (num1 + num2) * 10;Evaluation:
(1 + 2) * 10
= 3 * 10
= 30
Output: 30
Parentheses have the highest priority, so the addition is performed first.
- C++ follows BODMAS/PEMDAS rules.
- Multiplication and division have higher precedence than addition and subtraction.
- Use parentheses
()whenever you want a specific part of the expression to be evaluated first. -
💡 Tip: If you're ever unsure about the order of evaluation, use parentheses
(). They make the code easier to read and prevent logical mistakes.
Structure of any function- return_type function_name(parameters)
-
void is also a return type , it just that it is a special return type where function returns nothing.
-
every function has a return type int add() -> returns an integer | return type = int double area() -> returns a decimal | return type = doule void add() -> returns nothing | return type = void
-
void return also have parameters like other return types-
A. Void function without parameters- void greet() { cout << "Hello"; } Returns nothing, Takes nothing.
B. Void function WITH parameters- void add(int a, int b) { cout << a + b; }
Notice now, (int a, int b) are paramters Now you call- add(10, 20); Output- 30
- int function with & without parameters
A. No params- int getNumber() { return 5; }
No parameters, Returns an integer.
B. With params- int add(int a, int b) { return a + b; }
It takes both- ✔ takes parameters ✔ returns a value
void add2nos(int num1, int num2)
- Now the parameter list isn't empty anymore—the function receives num1 and num2 from whoever calls it.
For example: add2nos(10, 20);
Here: int num1, int num2 are parameters (they're defined in the function). 10, 20 are arguments (the actual values you pass when calling the function).
#include <iostream>
using namespace std;
void first() {
cout << "1";
}
int second() {
cout << "2";
return 3;
}
int main() {
cout << "A";
first();
cout << second();
cout << "B";
first();
cout << second();
cout << "C";
return 0;
}
Output - A123B123C
#include <iostream>
using namespace std;
int fun() {
cout << "A";
return 5;
}
int main() {
cout << fun() + fun();
return 0;
}
Output- AA10
Why its not A5A5
- when we write, return 5 the function is giving 5 back to the caller. It is not displaying 5.
- Only this line prints 5 (or any returned value): cout << fun();
- In your program, you wrote: cout << fun() + fun();
- So the two returned values are added first, and then the result is printed.
cout prints. return sends a value back.
Problem stat- Write a C++ program to create a void function that accepts two integers and prints their sum. Use a for loop to call the function multiple times and demonstrate code reusability all using void
#include <iostream>
using namespace std;
void add2nos() {
int num1, num2;
cout << "Enter 2 numbers: ";
cin >> num1 >> num2;
cout << "The sum of 2 numbers is: " << num1 + num2 << endl;
}
int main() {
for (int i = 1; i < 3; i++) {
add2nos();
}
return 0;
}
Problem stat- Write a C++ program to create a void function that accepts two integers and prints their sum. Ask the user how many times the function should be executed, then use a for loop to call the function that many times. Demonstrate code reusability using a void function.
#include <iostream>
using namespace std;
void add2nos() {
int num1, num2;
cout << "Enter 2 numbers: ";
cin >> num1 >> num2;
cout << "The Sum of 2 numbers is: " << num1 + num2 << endl;
}
int main() {
int n;
cout << "How many times do you want to add two numbers? ";
cin >> n;
for (int i=0; i < n; i++) {
add2nos();
}
return 0;
}
Enhancements-
instead of How many times do you want to add two numbers? we can write: How many times do you want to perform the addition? OR How many pairs of numbers do you want to add?
What if someone enters -5 ? Your program simply ends. It isn't wrong, but it isn't very user-friendly. A nicer program would validate the input.
if (n <= 0) { cout << "Please enter a positive number."; } else { for (int i = 0; i < n; i++) { add2nos(); } }
code is- int main() {
int n; cout << "How many times do you want to add two numbers? "; cin >> n;
if (n <= 0){ cout << "Enter a positive number!"; }
else {
for (int i=0; i < n; i++) { add2nos(); }
} return 0;
}
void add2nos() {
int num1, num2;
cout << "Enter 2 numbers: ";
cin >> num1 >> num2;
cout << "The Sum of 2 numbers is: " << num1 + num2 << endl;
}
int main() {
int n;
cout << "How many times do you want to add two numbers? ";
cin >> n;
if (n <= 0){
cout << "Enter a positive number!";
}
else {
for (int i=0; i < n; i++) {
cout << "Pair " << i + 1 << endl; // Pair Counter
add2nos();
}
}
return 0;
}
- Right now add2nos() does input, calculation & output
- Later, in larger programs, it's often better to separate responsibilities. For example: One function reads input. One function calculates. One function prints.
void inputNumbers(int &num1, int &num2) { cout << "Enter 2 numbers: "; cin >> num1 >> num2; }
int calculateSum(int num1, int num2) { return num1 + num2; }
void printSum(int sum) { cout << "The sum is: " << sum << endl; }
3 function does seperate things- input calculate & return print
Actual Code for SRP here-
#include <iostream>
using namespace std;
// Function 1: Read input
void inputNumbers(int &num1, int &num2) {
cout << "Enter 2 numbers: ";
cin >> num1 >> num2;
}
// Function 2: Calculate the sum
int calculateSum(int num1, int num2) {
return num1 + num2;
}
// Function 3: Print the result
void printSum(int sum) {
cout << "The sum is: " << sum << endl;
}
int main() {
int num1, num2;
int sum;
inputNumbers(num1, num2);
sum = calculateSum(num1, num2);
printSum(sum);
return 0;
}
Problem Statement- write a program that accepts numbers and print summation of 2 numbers & multiply the sum with 10 and print it. Store the sum in a variable
int add(int a, int b) {
return a + b;
}
int main() {
int num1, num2;
cout << "Enter two numbers: ";
cin >> num1 >> num2;
int result = add(num1, num2); // cout << "The sum is: " << add(num1, num2) << endl; - we can call the funtion inside cout also
cout << result << endl;
cout << result * 10 << endl;
return 0;
}
Modifications- storing sum & multiplied value into different variables
int add(int a, int b) {
return a + b;
}
int main() {
int num1, num2;
cout << "Enter 2 numbers: ";
cin >> num1 >> num2;
int sum = add(num1, num2);
int result = sum * 10;
cout << "The sum of 2 numbers is: " << sum << endl;
cout << "Multiplying sum with 10: " << result << endl;
return 0;
}
Problem Statement- write a program that accepts numbers of 10 & 20 and print summation of 2 numbers and their multiplication with 20 and we store the sum into variable
int add(int a, int b) {
return a + b;
}
int main() {
int num1 = 10;
int num2 = 20;
int result = add(num1, num2); // int result = add(10, 20); // cout << "The sum of " << num1 << " and " << num2 << " is: " << add(num1, num2) << endl;
cout << result << endl; // cout << "The sum of " << num1 << " and " << num2 << " is: " << add(num1, num2) << endl;
cout << result * 20;
return 0;
}
Paramterized functions-
A parameterized function is a function that accepts values as input through parameters, allowing the same function to work with different data.
The Essence of a parameterized function: same function, different inputs, different outputs
Why we need parameterized function?
- becasue we don't want any modifications later, and as our input gets changed over time, the output should come in desired results
- for example-> void greet() { cout << "Hello Sarbesh"; }
- Output-> Hello Sarbesh
- now tomorrow if we want Hello John, Hello Anu, Hello Cutie then? will we create different functions for them like greetAnu() or greetCutie() ❌
- Instead, we make the function accepts Input
Parameters-> void greet(string name) here: Function Name -> greet Parameter -> name Now the function can work with any name.
() -> Empty parantheses -> meaning no info passing into the function -> Means the function needs nothing from the outside (params) -> from now on we will put some info
for eg: void greet(string name) -> parantheses are not empty, () contain string name, means Before I can do my work, you must give me one string
// paramterized functions using void Problem stat- Write a C++ program that creates a parameterized void function to greet a person by name. Call the function multiple times with different names.
#include <iostream>
using namespace std;
void greet(string name) {
cout << "Hello " << name << endl;
}
int main() {
greet("Sarbesh");
greet("Anu");
return 0;
}
Old code when no endl is given inside void so the name was priniting on same line and i cleverly wrote cout << endl; in code line 3042
void greet(string name) {
cout << "Hello " << name;
}
int main() {
greet("Sarbesh");
cout << endl;
greet("Anu");
return 0;
}
void greet(string name) { cout << "Hello " << name << endl; }
-
greet is just a function name
-
we could also write void hello(string name) or void banana(string name) , the compiler dosen't care
-
string name is a parameter
-
Means- This function expects a string. When someone calls me, I'll store that string inside a variable called name.
-
think of it like an empty box, before calling name = ? , after calling greet("Sarbesh") the CPU does name = "Sarbesh"
-
now the fucntion becomes - cout << "Hello " << "Sarbesh" << endl;
-
Output - Hello Sarbesh
-
why hello remains same because it is fixed and name is a variable and so it gets changed everytime cout << "Hello " << name;
-
and coming why we choose name, we can actually choose anything over here like person or xyz
-
like void greet(string person) cout << "Hello " << person;
-
the compiler cares about consistency
-
like for example-
void greet(string banana) { cout << "Hello " << banana; } greet("Sarbesh");
Output- Hello Sarbesh
Why is it called a parameterized function?
- Because the function accepts a parameter.
void greet(string name)
Here, string name is called the parameter. When calling- greet("Sarbesh"); then "Sarbesh" is called the argument.
So:
void greet(string name)
name → parameter
greet("Sarbesh");
"Sarbesh" → argument
There is only one variable called name But we got: Hello Sarbesh Hello Anu
Why? Because every time you call the function, greet(...) the parameter receives a new value.
One thing I'd like you to notice, Earlier you asked:
"Can I name num1 and num2 as a and b?" Exactly the same rule applies here. This:
- void greet(string name) could also be
- void greet(string x) or
- void greet(string person) or
- void greet(string studentName)
All are valid. For example:
void greet(string studentName) { cout << "Hello " << studentName << endl; }
works exactly the same. The compiler only cares about the type (string), not the variable name.
void greet(string name)
means: Function name = greet Parameter type = string Parameter variable = name Whatever string is passed during function call, gets stored in name.
- Current version:
void greet(string name) greet("Sarbesh"); // when you call
C++ creates a copy Think: Original: "Sarbesh"
Copy: "Sarbesh"
Function uses the copy
- Reference version: void greet(string& name)
Now no copy is created, Function directly uses the original string, Faster.
later we will use- void greet(const string& name)
Meaning: Don't make a copy Use the original string Don't allow modifications
void show(int x) { cout << x; } show(25); // Output- 25
void show(int x) { -> parameter head cout << x + 5; -> parameter body } show(10); // Output- 15 -> show(10) is a an argument
int add(int a, int b) { return a + b; } cout << add(2, 3); // Output- 5
void printSalary(double salary) { cout << salary; } printSalary(12345.67); // Output- 12345.67
- void student(string name, int age, char grade) { cout << name << " " << age << " " << grade; } student("Sarbesh", 22, 'A'); // Output- Sarbesh 22 A
-
Remember the Argument should match the parameter type
-
when we see: void greet(string name) Function Name : greet Input Type : string Variable Name : name
-
when we se: void add(int a, int b) Function Name : add Input Type : int, int Variable Name : a, b
-
The pattern is always->
return_type function_name(parameter_type parameter_name)
- code 1-
void greet(string name) {
cout << "Hello " << name << endl;
}
int main() {
string person = "Rahul";
greet(person);
return 0;
}
Output- Hello Rahul
- Code 2-
void greet(string name) {
cout << "Hello " << name << endl;
}
int main() {
string person = "Rahul";
greet("person");
return 0;
}
Output- Hello person
There is a difference between:
greet(person);
and
greet("person");
Suppose:
string person = "Rahul";
- greet(person); → passes the VALUE stored in the variable (
"Rahul"). - greet("person"); → passes the literal string
"person".
Output: greet(person); → Hello Rahul greet("person"); → Hello person
Variables pass their stored value, not their variable name.
- Code 3-
void greet(string name) {
cout << "Hello " << name << endl;
}
int main() {
string x = "Sarbesh";
string y = "Anu";
greet("x");
greet(y);
return 0;
}
Output - Hello x Hello Anu
Notice: greet(x); means "Go look inside the variable."
Whereas greet("x"); means "Don't look anywhere. Just use the word x."
Images for more example / clarification-
code-
void greet(string name) {
cout << "Hello " << name << endl;
}
int main() {
string person = "Rahul";
person = "Amit";
greet(person);
return 0;
}
Output- Hello Amit
Explanation-
person = "Amit"; This does not create another variable. It changes the existing variable.
Where is Rahul? Gone. It has been replaced.
Thing to remember- Ealier i was thinking the varibale value will not get updated cuz of pass by value or reference... but NO, Pass-by-value or reference only starts after the function is called.
code-
void greet(string name) {
name = "Rohit";
cout << name << endl;
}
int main() {
string person = "Amit";
greet(person);
cout << person << endl;
}
Output- Rohit Amit
Explanation-
-
Before function call: main() person = Amit
-
After function call:
greet(person); A copy is made.
main() person = Amit
greet() name = Amit
- Inside function:
name = "Rohit";
Memory now becomes:-
main() person = Amit
greet() name = Rohit
cout << name prints Rohit
The function changed name NOT person Those are two different variables.
-
Function ends. Everything inside greet() disappears. name is destroyed.
-
Back to main() The only variable left is person = Amit cout << person; // Amit is printed in last part of output
greet() prints Rohit. main() prints Amit.
code-
void greet(string name) {
name = "Charlie";
cout << "Inside function: " << name << endl;
}
int main() {
string person = "Alice";
greet(person);
cout << "Inside main: " << person << endl;
return 0;
}
Output- Inside function Charlie Inside main Alice
Insights-
- person contains Alice before function call
- name receives Alice
- inside function:
Inside function Charlie got printed cuz name = "Charlie"; changed only name - inside main() after function call: Inside main: Alice cuz person never changed
void greet(string name) { name = "Charlie"; cout << "Inside function: " << name << endl; }
int main() { string name = "Alice"; greet(name); cout << "Inside main: " << name << endl;
return 0; }
Output- Same output of previous one:
Charlie Alice
Observation-
- person varibale got changed to name variable inside main()
- so inside void and isnide main both have variable named- name
- but both name are not same varibales, they look identical but r not same
- Only the local copy changed inside void greet() it became Alice to Charlie
- Function ends, everything inside void greet() disappears and only one variable left i.e Alice and that is printed as 2nd output inside main()
- So no colliding of varibale value or variable name happens
This is why changing one doesn't affect the other. main()
name = Alice
↓
copy
↓
greet()
name = Alice
↓
change
↓
greet()
name = Charlie
↓
destroy
↓
main()
name = Alice
The original never changed.
Note-
- Variables with the same name can exist in different functions.
- They do NOT refer to the same variable.
- Each function has its own local scope.
- When a function is called by value, a copy of the argument is created.
- Changing the parameter inside the function does NOT change the original variable in
main(). - When the function ends, only its local variables are destroyed. Variables in
main()remain unchanged.
Paramerterized functions with returning functions-
lets's take an example why we need return instead of void->
void greet(string name) { cout << "Hello " << name; } greet("Sarbesh"); // Call
Output-> Hello Sarbesh
Can we do this? -> int x = greet("Sarbesh"); ofc NO ❌ cuz greet() returns nothing.
- Returning Function->
int add(int a, int b) {
return a + b;
}
- here this function promises to return an integer
- Suppose-> add(10, 20);
- CPU enters a = 10 and b = 20
- then return (a + b); becomes return 30; Functions sends 30 back to whoever called it
Where does this 30 goes?
- Option1:- Print immediately
cout << add(10,20);
Execution: add(10,20) ↓ returns 30 ↓ cout << add(10,20) // prints 30 // Output- 30
-
Option2:- Store it
int sum = add(10,20); // Now sum = 30 cout << sum; // Output- 30
-
Option3:- Reuse it
int sum = add(10,20); cout << sum * 10;
Output- 300
Notice that Function code unchanged, but we did something completely different with the result. This is why returning functions are powerful.
string getName() {
return "Sarbesh";
}
cout << getName(); // Output- Sarbesh
In void output happens inside function AND in returning output happens wherever we choose
Void-> void add(int a, int b) { cout << a + b; }
Return-> int add(int a, int b) { return a + b; }
-
cout << add(10,20);
-
int result = add(10,20);
-
if(add(10,20) > 25)
all works, whenever we want output we can do
prblm stat- Write a parameterized function that accepts two integers and returns their sum. Call the function from main() and print the returned value.
#include <iostream>
using namespace std;
int sum(int a, int b) {
return a + b;
}
int main() {
int res = sum(4,5) + 10; // + 10 is not required as per prblm statement but just for concept =, it shows functions can be nested inside expressions
cout << res << endl;
return 0;
}
output- 19 process- returns 9 -> 9+10 = 19
Time complexity- return a + b; takes constant time time complexity- O(1)
int sumOfTwoNumbers(int a, int b)
{
cout << "Inside function" << endl;
return a + b;
}
int main()
{
cout << "A" << endl;
cout << sumOfTwoNumbers(2, 3) << endl;
cout << "B" << endl;
return 0;
}
Output- A Inside function 5 B
Note- "Inside function" is printed by the cout inside the function, while 5 is printed by the cout inside main(), right
int add(int a, int b)
{
cout << "Function Starts" << endl;
return a + b;
}
int main()
{
cout << "A" << endl;
int x = add(2, 3);
cout << "B" << endl;
cout << x << endl;
cout << "C" << endl;
return 0;
}
Output- A Function starts B 5 C
N/B:- at exactly int x = add(2,3) function call happens B is printed after the function finishes cuz cout << x is after cout << "B"
int add(int a, int b)
{
cout << "Inside add()" << endl;
return a + b;
}
int main()
{
cout << "Start" << endl;
cout << add(2, 3) + add(4, 5) << endl;
cout << "End" << endl;
return 0;
}
Output- Start Inside add() Inside add () 14 End
Note-
-
2 times function call so, add() executes twice
-
can you guess which function executed first?
Ans- Ofc NO cout << add(2, 3) + add(4, 5); the order in which the two function calls are evaluated is not guaranteed by the language. That means the compiler is allowed to choose. One compiler might do add(2,3) and then add(4,5) -> 5 + 9 another compiler might do add(4,5) then add(2,3) -> 9 + 5 both produce 14 because addition is commutative.
-
If we modify the code slightly we can predict which function got executed first
int add(int a, int b) { cout << "Adding " << a << " and " << b << endl; return a + b; }
Now the output might be Compiler A: Start Adding 2 and 3 Adding 4 and 5 14 End
Compiler B: Start Adding 4 and 5 Adding 2 and 3 14 End
int add(int a, int b)
{
cout << "Function\n";
return a + b;
}
int main()
{
int x = add(2,3);
cout << x + 10 << endl;
}
Output- Function 15
Remember--- A function can do two independent things:
- It can print something using cout.
- It can return a value using return.
These are completely different operations.
- cout displays output on the screen.
- return sends a value back to the caller.
// Parameterized Functions using return // Problem Statement- Write a function that returns addition of 2 numbers
#include <iostream>
using namespace std;
int sumoftownos(int num1, int num2) {
return num1 + num2;
}
int main() {
int res = sumoftownos(10, 20); // cout << sumoftownos(10, 20); can also be used but we used this if we need sumoftownos multiple times
cout << res;
return 0;
}
int sumOfTwoNumbers(int num1, int num2) num1 and num2 are parameters
sumOfTwoNumbers(10, 20); 10 and 20 are arugments
int add(int a, int b) {
return a + b;
}
int main() {
int x = add(3, 4);
int y = add(x, 10);
cout << x << "\n" << y;
return 0;
}
can you see that we wrote x in ( int y = add(x, 10); ) we are passing 7 and not directly wrote 7 as x contains 7 only
Prblm stat- Write a parameterized function that returns the sum of two integers. Call the function multiple times with different values, store each returned value in separate variables, and print the results.
#include <iostream>
using namespace std;
int add(int a, int b) {
return a + b;
}
int main() {
int x = add(3,4);
int y = add(x,10);
int z = add(y,x);
int s = add(add(x,y),10);
cout << x << endl << y << endl << z << endl << s;
}
#include <iostream>
using namespace std;
void change(int x) {
x = 100;
cout << x << endl;
}
int main() {
int num = 10;
change(num);
cout << num;
}
Output-
100 10
prblm stat- Write a void function that accepts an integer by value, changes its value inside the function, and demonstrate that the original variable in main() remains unchanged.
#include <iostream>
using namespace std;
void change(int x) {
cout << "Inside before: " << x << endl;
x = 100;
cout << "Inside after : " << x << endl;
}
int main() {
int num = 10;
cout << "Before function: " << num << endl;
change(num);
cout << "After function: " << num << endl;
return 0;
}
Output- Before function 10 Inside before 10 Inside after 100 After function 10
void change(int x)
{
x = x + 5;
cout << x << endl;
}
int main()
{
int num = 10;
change(num);
cout << num << endl;
}
Output- 15 10
Prblm stat- Write a returning function that accepts an integer by value, changes its local copy, returns the modified value, and show that the original variable in main() is unaffected.
#include <iostream>
using namespace std;
int change(int x) {
x = 100;
return x;
}
int main() {
int num = 10;
cout << change(num) << endl;
cout << num;
return 0;
}
Output- 100 10
Prblm stat- Write a returning function that accepts an integer by value, changes its local copy, returns the modified value, and show that the original variable in main() is unaffected.
#include <iostream>
using namespace std;
int change(int x) {
x = 100;
return x;
}
int main() {
int num = 10;
int result = change(num);
cout << "Returned value: " << result << endl;
cout << "Original value: " << num;
return 0;
}
Output- Returned value: 100 Original value: 10
#include <iostream>
using namespace std;
int change(int x) {
x = 100;
return x;
}
int main() {
int num = 10; // Call started from here
cout << change(num) << endl;
cout << num;
return 0;
}
Output- 100 10
Initially: num = 10
Call: change(num);
Pass by value creates a copy: num = 10 copy x = 10
Inside function: x = 100;
Now: copy x = 100; num = 10;
Return: return x; returns 100
Ouput: 100 10
Returned value = 100 Original variable = 10
#include <iostream>
using namespace std;
void change(int &x) { // here addition of & makes it reference
x = 100;
}
int main() {
int num = 10;
change(num);
cout << num;
}
Output- 100
- No copy is created, instead x becomes another name for num , so both point to same variable
Prblm stat- Write a void function that accepts an integer using pass by reference. Modify the value inside the function and print the value in main() after the function call to demonstrate that changes made through the reference affect the original variable.
#include <iostream>
using namespace std;
void change(int &x) {
x = 100;
cout << x << endl;
}
int main() {
int num = 10;
change(num);
cout << num;
}
Output- 100 100
note-
- & is not a pointer
- here, & means create another name for the same variable
#include <iostream>
using namespace std;
void change(int &x)
{
cout << "Inside before: " << x << endl;
x = x + 20;
cout << "Inside after: " << x << endl;
}
int main()
{
int num = 15;
cout << "Main before: " << num << endl;
change(num);
cout << "Main after: " << num << endl;
return 0;
}
Output- Main before 15 inside before 15 Inside after 35 Main after 35
explanation of & in mnetal mode- 
Write a function that accepts an integer by reference, changes its value inside the function, returns the modified value, and demonstrate that the original variable in main() is also changed.
#include <iostream>
using namespace std;
int change(int &x) {
x = 100;
return x;
}
int main() {
int num = 10;
cout << change(num) << endl;
cout << num;
return 0;
}
Output- 100 100
Output- 100 100
Initally: num = 10
Call: change(num);
Reference: x refers to num
inside function: x = 100;
Since x and num are same varibales- num = 100
Return: return x; returns 100
Output- 100 100
Returned value = 100 Original variable = 100
Write a function that accepts an integer by reference, changes its value inside the function, returns the modified value, and demonstrate that the original variable in main() is also changed.
#include <iostream>
using namespace std;
int change(int &x) {
x = 100;
return x;
}
int main() {
int num = 10;
int updatedValue = change(num);
cout << "Returned value: " << updatedValue << endl;
cout << "Return after function: " << num;
return 0;
}
#include <iostream>
using namespace std;
int update(int &x)
{
x = x + 5;
return x;
}
int main()
{
int num = 10;
int result = update(num);
cout << num << endl;
cout << result << endl;
}
Output- 15 15
Mental model-
int update (int &x) │ │ │ │ │ └── How the parameter is received │ └────────────────── Function name └──────────────────────────────── What the function returns
when we pass arr to a function
- C++ does not copy the entire array.
- instead, it passes the address of the first element.
One thing you may not know yet:
int
↓
copy
int& ↓ reference
int arr[] ↓ Array parameters automatically behave like pointers.
-
The function receives the address of the first element, not a copy of the entire array.
-
Therefore changes inside the function affect the original array.
-
Arrays decay into pointers when passed to functions. So the function receives the address of the first element.
-
in array we dont have to write & because it is passing the address of the first element and technically it is pointer
int arr[] -> pointer
Quick test->
Without running:
void fun(int arr[]) {
arr[2] = 50;
}
int main() {
int arr[] = {1, 2, 3, 4, 5};
fun(arr);
cout << arr[2];
}
Output-> 50
Prblm Stat- Write a void function that accepts an integer array, changes the first element of the array and demonstrate that the original array in main() is also modified.
#include <iostream>
using namespace std;
void explainPassByValueAndReference(int arr[]) { // instead of arr[] we can write *arr as for arrays we can write * before it
arr[0] = 100; // for array we don't hv to write & to pass the value cuz its already taking a reference
}
int main() {
int arr[] = {6, 7, 8, 1, 2};
explainPassByValueAndReference(arr); // It takes memory locations and not copy the data
cout << arr[0]; // Passing the memory locations and not copying
// the function name is so long, we can make it modifyArray() or updateArray()
return 0;
}
Output- 100
Note-
void fun(int arr[]) = void fun(int *arr)
In main: arr
+----+----+----+----+----+ | 6 | 7 | 8 | 1 | 2 | +----+----+----+----+----+ ^ address
When you call: explainPassByValueAndReference(arr);
the function recieves: arr -> address of first element
So inside the function, arr points to the same array. It is not another array.
when we write: arr[0] = 100;
we are changing: +----+----+----+----+----+ |100 | 7 | 8 | 1 | 2 | +----+----+----+----+----+
we are chaning the the same array that exists in main(). So after the function returns, main() still has 100
why & not needed?
int arr[] is treated as a pointer to the first element when passed to a function. That pointer already refers to the original array. So another & isn't needed.
Explanation-
explainPassByValueAndReference(arr); does not pass the entire array in c++ instead it passes Address of first element
Think: arr Index Value 0 6 1 7 2 8 3 1 4 2
Suppose arr[0] lives at address 1000 then, explainPassByValueAndReference(arr); = explainPassByValueAndReference(1000);
Arrays decay into pointers when passed to functions. So the function receives the address of the first element.
in array we dont have to write & because it is passing the address of the first element and technically it is pointer
int arr[] -> pointer
we use comma in cpp for different purposes:
- Comma separating variables ✅ int a = 5, b = 10;
or
int a, b, c;
means, Create two variables. Here comma is used as a seperator & not operator
- Comma separating function arguments ✅ add(2, 3);
means, Pass two arguments. comma is seperator
- Comma operator
⚠️ (5,10)
means, Evaluate 5, ignore it, return 10. it's operator not seperator
- the compiler is no longer looking for a separator. It is looking for one expression inside the brackets.
- arr[ ??? ] needs one value
- arr[0,1] becomes arr[1]
What is an expression inside []? Anything inside square brackets [] is called the index expression.
arr[0] is simply 0
but it dosen't have to be a number eg: int i = 2; cout << arr[i];
here the expression is i which evaluates to 2
So anything that finally becomes an integer can go inside [].
[] -> Index {} -> Number
int arr[] = {5,8,9};
0th index is 5, 1st index is 8, 2nd index is 9
arr[1] -> Give me the element stored at index 1. which is 8
is it all the time we have integers inside [] It is not only integer numbers written directly. It is anything that evaluates to an integer value.
All these are valid:- ✔
arr[0] // integer literal arr[5]
int i = 2; arr[i] // variable containing an integer
arr[1 + 2] // expression -> becomes 3
arr[i + 1] // expression -> becomes an integer
arr[x * 2] // expression -> becomes an integer
All these are not valid:- ❌
arr[3.5] // double (not an integer index) arr["hello"] // string arr[true] // technically converts to 1 in C++, but don't use this
when we write: int i = 2; arr[i];
- it means Go to the index stored in i
Prblm stat- Write a void function that accepts an integer array, modifies the first two elements of the array, and demonstrate that the changes are reflected in the original array in main().
#include <iostream>
using namespace std;
void updateArray(int arr[]) {
arr[0] = 100;
arr[1] = 200;
}
int main() {
int arr[] = {6, 7, 8, 1, 2};
updateArray(arr);
cout << arr[0] << endl;
cout << arr[1] << endl;
cout << arr[2] << " " << arr[3];
return 0;
}
Output-
100 200 8 1
for loop-
instead of: cout << arr[0] << endl; cout << arr[1] << endl; cout << arr[2] << " " << arr[3];
we can use:
for(int i = 0; i < 5; i++) { cout << arr[i] << " "; }
Output- 100 200 8 1 2
Phenomenon- when we call: updateArray(arr);
function parameter int arr[] does not recieve a copy of the array. Instead, it refers to the same array that exists in main().
Before the call: main() arr
Index 0 1 2 3 4 Value 6 7 8 1 2
Function is called: updateArray(arr)
Inside the function: arr
Index 0 1 2 3 4 Value 6 7 8 1 2
There is not another array. Both main() and updateArray() are looking at the same memory.
So when this happens arr[0] = 100; the original array changes immediately. then, arr[1] = 200; changes the 2nd element too
When the function finishes, nothing is copied back because nothing was copied in the first place.
#include <iostream>
using namespace std;
void changeArray(int arr[], int size)
{
for(int i = 0; i < size; i++)
{
arr[i] = 100;
}
}
int main()
{
int arr[] = {5,8,3};
changeArray(arr,3);
for(int i=0;i<3;i++)
{
cout << arr[i] << " ";
}
return 0;
}
Output- 100 100 100
Why arr[i] = 100; ???
Suppose int arr[] = {5,8,3};
Index 0 1 2 Value 5 8 3
Loop: for(int i=0;i<3;i++) { arr[i]=100; }
1st iteration: i=0
Compiler sees,
arr[i] which becomes arr[0]
So,
arr[0]=100;
Array becomes: 100 8 3
2nd iteration: i=1
arr[1]=100; Array becomes: 100 100 3
3rd iteration: i=2
arr[2]=100; Array becomes: 100 100 100
That's why we write arr[i] because i changes, so different elements are modified.
Prblm stat- Write a function that accepts an integer array, modifies the first two elements of the array, returns the first modified element, and demonstrate that both the returned value and the original array in main() are updated.
#include <iostream>
using namespace std;
int updateArray(int arr[]) {
arr[0] = 100;
arr[1] = 200;
return arr[0];
}
int main() {
int arr[] = {1,2,3,4,5};
int x = updateArray(arr);
cout << "Returned value: " << x << endl;
cout << arr[0] << "\n" << arr[1] << "\n" << arr[2] << "\n" << arr[3] << "\n" << arr[4];
return 0;
}
Output- Returned value: 100 100 200 3 4 5
- Notice something important.
- The return statement did not modify the array. The array had already changed before the return happened. The return simply sends back one value.
little bit changing how output is represented-
instead of, cout << arr[0] << "\n" << arr[1] << "\n" << arr[2] << "\n" << arr[3] << "\n" << arr[4]; Output- 100 200 3 4 5
we can do, cout << arr[0] << " " << arr[1] << " "<< arr[2] << " " << arr[3] << " " << arr[4]; Output- 100 100 200 3 4 5
- Without parentheses in function, you're referring to the function itself, not calling it.
- A function name can decay to a function pointer. When streamed with cout, the output is implementation-dependent, and in your environment it ended up displaying 1. That's not the function's return value.
- updateArray means function
- updateArray(arr) means run the function
instead of cout << x where x = updateArrat(arr) = function call i wrote cout << updateArray ❌
#include <iostream>
using namespace std;
int updateArray(int arr[])
{
arr[0]++;
return arr[0];
}
int main()
{
int arr[] = {5, 6, 7};
cout << updateArray(arr) << endl;
cout << updateArray(arr) << endl;
cout << arr[0];
}
Output- 6 7 7
- A return statement can return any valid expression, not just a variable.
- we can do anything with return
- we can return:-
- return arr[0] + arr[1];
- return arr[2] * arr[4];
- return arr[0] + arr[1] + arr[2] + arr[3] + arr[4];
- return arr[0] * 10;
- The return statement is simply saying: "Evaluate this expression and give me its result."
"Evaluate this expression and give me its result."
It doesn't care whether the expression is:
- one variable,
- one array element,
- a calculation,
- a function call,
- or something much more complex.
int updateArray(int arr[])
{
arr[0] = arr[1] + arr[2];
return arr[0] + arr[1];
}
int main()
{
int arr[] = {2,4,6};
cout << updateArray(arr) << endl;
cout << arr[0] << " " << arr[1] << " " << arr[2];
}
output- 14 10 4 6
int update(int arr[])
{
arr[1] = arr[0] + arr[2];
arr[2] = arr[1] + 5;
return arr[2];
}
int main()
{
int arr[] = {3, 7, 2};
cout << update(arr) << endl;
cout << arr[0] << " "
<< arr[1] << " "
<< arr[2];
}
Output- Output- 10 3 5 10
old code-
#include <iostream>
using namespace std;
int updateArray(int arr[]) {
arr[0] = 100;
arr[1] = 200;
return arr[0] + arr[1] + arr[2] + arr[3] + arr[4]; // we will automate this and we dont hv to write return arr[0] +arr[2] +... manually
}
int main() {
int arr[] = {1,2,3,4,5};
int x = updateArray(arr);
cout << "Returned value: " << x << endl;
cout << arr[0] << "\n" << arr[1] << "\n" << arr[2] << "\n" << arr[3] << "\n" << arr[4]; // we are also automating this using loop
return 0;
}
Loop version-
#include <iostream>
using namespace std;
int updateArray(int arr[]) {
arr[0] = 100;
arr[1] = 200;
int sum = 0;
for (int i = 0; i < 5; i++) {
sum = sum + arr[i];
}
return sum;
}
int main() {
int arr[] = {1,2,3,4,5};
int x = updateArray(arr);
cout << "Returned Sum: " << x << endl;
for (int i = 0; i < 5; i++) {
cout << arr[i] << endl;
}
return 0;
}
Output- Returned Sum: 312 100 200 3 4 5
Explanation-
1, 2, 3, 4, 5 becomes 100, 200, 3, 4, 5 after arr[0] = 100; arr[1] = 200;
Now the loop starts:
First iteration: i = 0 sum = 0 + 100 = 100
Second iteration: i = 1 sum = 100 + 200 = 300
Third iteration: i = 2 sum = 300 + 3 = 303
Fourth iteration: i = 3 sum = 303 + 4 = 307
Fifth iteration: i = 4 sum = 307 + 5 = 312
Loop ends. Function returns 312 312 gets printed because cout << function name = cout << x cout << "Returned Sum" << x
Modifications- Noticed we wrote i < 5 everywhere but if the array size increases in future we will hv to update 5 from everywhere better we use array size which will pass the size to function
code-
#include <iostream>
using namespace std;
int updateArray(int arr[], int size) {
arr[0] = 100;
arr[1] = 200;
int sum = 0;
for (int i = 0; i < size; i++) {
sum = sum + arr[i];
}
return sum;
}
int main() {
int arr[] = {1,2,3,4,5};
int size = 5;
int x = updateArray(arr, size);
cout << "Returned Sum: " << x << endl;
for (int i = 0; i < size; i++) {
cout << arr[i] << endl;
}
return 0;
}
Problem Statement: Write a void function that accepts an integer array, uses a loop to update the first three elements of the array to 100, demonstrates that the original array inmain() is modified, and prints the updated array.
#include <iostream>
using namespace std;
void updateArray(int arr[])
{
for(int i = 0; i < 3; i++)
{
arr[i] = 100;
}
}
int main()
{
int arr[] = {6,7,8,1,2};
updateArray(arr);
for(int i = 0; i < 5; i++)
{
cout << arr[i] << " ";
}
return 0;
}
Output- 100 100 100 1 2
Understanding- Initially
Index : 0 1 2 3 4 Value : 6 7 8 1 2
First iteration: i = 0 Compiler executes: arr[i] = 100;
becomes arr[0] = 100;
Array: 100 7 8 1 2
Second iteration: i = 1 Compiler executes: arr[1] = 100;
Array: 100 100 8 1 2
Third iteration: i = 2 Compiler executes: arr[2] = 100;
Array: 100 100 100 1 2
Loop ends because i = 3 and 3 < 3 is false.
Why write arr[i] = 100;
instead of
arr[0] = 100; arr[1] = 100; arr[2] = 100;
Because the loop changes which index is being accessed each time.
Think of i as a moving finger. arr[i] = 100; never changes. Only the value of i changes.
Prblm stat- Every element should become twice of its original value like array 1,2,3,4,5 should become array 2,4,6,8,10
1 2 3 4 5 -> 2 4 6 8 10
Prblm stat- Write a void function named doubleArray() that accepts an integer array. Use a for loop to visit every element of the array and double its value. Demonstrate that the original array in main() is modified by printing the updated array after calling the function.
#include <iostream>
using namespace std;
void doubleArray(int arr[]) {
for(int i = 0; i < 5; i++) {
arr[i] = arr[i] * 2;
}
}
int main() {
int arr[] = {1,2,3,4,5};
doubleArray(arr);
for(int i = 0; i < 5; i++) {
cout << arr[i] << endl;
}
return 0;
}
Output- 2 4 6 8 10
Phenomenon-
What is the purpose of a for loop? A for loop is used when we want to perform the same operation repeatedly.
Instead of writing: arr[0] = arr[0] * 2; arr[1] = arr[1] * 2; arr[2] = arr[2] * 2; arr[3] = arr[3] * 2; arr[4] = arr[4] * 2;
we write: for(int i = 0; i < 5; i++) { arr[i] = arr[i] * 2; }
The loop performs the same operation on every element automatically.
The variable i is not the value inside the array. It is the index (position) of the element currently being visited.
i tells which box i am looking at
arr[i] means "Access the value stored at index i."
for eg: when i = 2; then arr[i] becomes arr[2] which is 3 in {1,2,3,4,5}
arr[i] tells what's the value inside the box.
Example- Think of an apartment building.
Apartment Number (Index)
0 1 2 3 4
Inside each apartment lives a number.
0 -> 1 1 -> 2 2 -> 3 3 -> 4 4 -> 5
i moves between apartments.
arr[i] is the person living inside.
i mistakenly wrote earlier- void doubleArray(int arr[]) { for (int i = 0; i < 5; i = i * 2) }
-
This will be an infinite loop because initially i = 0 then i=i2 becomes 0 = 02 = 0 so i = 0 forever , the condition is always true. So the loop never ends
-
when i wrote i = i * 2; I moved the apartment number. But the challenge wanted you to change the people living inside the apartments. So the correct line is arr[i] = arr[i] * 2;
-
the left side arr[i] tells the computer store the new value back into this position.
-
the right side arr[i] * 2 tells the computer Take the current value and multiply it by 2.
Dry Run-
Initially: Index : 0 1 2 3 4 Value : 1 2 3 4 5
Iteration 1: i = 0 arr[0] = 1 × 2
Array: 2 2 3 4 5
Iteration 2: i = 1
arr[1] = 2 × 2
Array: 2 4 3 4 5
iteration 3: i = 2
arr[2] = 3 × 2
Array: 2 4 6 4 5
Iteration 4: i = 3
arr[3] = 4 × 2
Array: 2 4 6 8 5
Iteration 5: i = 4
arr[4] = 5 × 2
Array: 2 4 6 8 10
Pattern-
-
The loop itself almost never changes. for(int i = 0; i < size; i++)
-
What changes is only the operation inside the loop. Eg:-
Double- arr[i] = arr[i] * 2;
Square- arr[i] = arr[i] * arr[i];
Make every element 100- arr[i] = 100;
Add 10- arr[i] = arr[i] + 10;
Subtract 5- arr[i] = arr[i] - 5;
- The loop is exactly the same. Only the operation changes.
- The only thing that changes is what you do to each element inside the loop.
whenever we write- for(int i = 0; i < size; i++)
ask yourself these two questions:
- Question 1:
What is i supposed to do? Answer: Visit each index.
- Question 2: What should happen to arr[i]?
Answer: Maybe
double it square it make it 100 print it sum it
The operation changes. The loop almost never changes.
Some pattern-
-
Make every element 100 for(int i = 0; i < size; i++) { arr[i] = 100; }
-
Double every element for(int i = 0; i < size; i++) { arr[i] = arr[i] * 2; }
-
Square every element for(int i = 0; i < size; i++) { arr[i] = arr[i] * arr[i]; }
Imagine the array is a row of lockers.
Locker Number (Index)
0 1 2 3 4
Inside the lockers are numbers: 0 → 1 1 → 2 2 → 3 3 → 4 4 → 5
-
The loop says: "Walk to every locker." That's the job of i++
-
Then, when you reach a locker, do something with what's inside. For this program, that "something" is:
arr[i] = arr[i] * 2;
So the loop is like a worker walking down the lockers: Walk to Locker 0 → Double the value
Walk to Locker 1 → Double the value
Walk to Locker 2 → Double the value
Walk to Locker 3 → Double the value
Walk to Locker 4 → Double the value
Problem Statement:
// Write a void function named transformArray() that accepts an integer array. // Use a for loop to visit every element of the array.
// If the element is even, multiply it by 2. // If the element is odd, multiply it by 3.
// Demonstrate that the original array in main() is modified // by printing the updated array after calling the function.
#include <iostream>
using namespace std;
void transformArray(int arr[]) {
for (int i = 0; i < 6; i++) {
if (arr[i] % 2 == 0) {
arr[i] = arr[i] * 2;
}
else {
arr[i] = arr[i] * 3;
}
}
}
int main() {
int arr[] = {2, 5, 8, 3, 10, 7};
transformArray(arr);
for (int i = 0; i < 6; i++) {
cout << arr[i] << " ";
}
return 0;
}
Output- 4 15 16 9 20 21
Pattern- many programmer follow this pattern-
for(...) { if(...) { // do something } else { // do something else } }
SOP- A small rule that will help you a lot
Whenever you solve an array problem, ask yourself these questions in order:
- Am I visiting every element? If yes →
for(...)
- Do I need to treat every element the same? If yes →
for(...) { // operation }
Example: arr[i] *= 2;
- Do I need to decide what to do based on each element? If yes →
for(...) { if(...) { // operation A } else { // operation B } }
Mental model- Here's a mental model that might stick
Imagine you're checking students entering a classroom.
Student 1 Student 2 Student 3 Student 4
The for loop is you walking past each student.
When you reach one student, you ask:
"Are you wearing a blue shirt?"
That's the if.
If the answer is yes, you hand them a blue card.
If the answer is no, you hand them a red card.
You don't walk around the whole classroom again just because one student was wearing blue.
#include <iostream>
using namespace std;
void transformArray(int arr[], int size) {
for (int i = 0; i < size; i++) {
if (arr[i] % 2 == 0) {
arr[i] = arr[i] * 2;
}
else {
arr[i] = arr[i] * 3;
}
}
}
int main() {
int arr[] = {2, 5, 8, 3, 10, 7};
int size = 6;
transformArray(arr, size);
for (int i = 0; i < size; i++) {
cout << arr[i] << " ";
}
return 0;
}
Output- 4 15 16 9 20 21
note- right now i wrote: int size = 6
later: int size = sizeof(arr) / sizeof(arr[0]);
Suppose: int arr[] = {2,5,8,3,10,7};
Each int usually takes 4 bytes. So, 6 elements × 4 bytes = 24 bytes
Therefore, sizeof(arr) returns 24 and, sizeof(arr[0]) returns 4
then, 24 / 4 = 6 So, size becomes 6 automatically.
Convert 5 10 15 20 to 25 100 225 400 array
#include <iostream>
using namespace std;
void transformArray(int arr[], int size) {
for (int i = 0; i < size; i++) {
arr[i] = arr[i] * arr[i];
}
}
int main() {
int arr[] = {5, 10, 15, 20};
int size = 4;
transformArray(arr, size);
for(int i = 0; i < size; i++) {
cout << arr[i] << " ";
}
return 0;
}
A for loop's job is to visit each element one by one. The real problem-solving happens inside the loop, where you decide what operation to perform on the current element (arr[i]).
That idea will stay true when you learn:
- Searching
- Sorting
- Strings
- Vectors
- Hashing
- Even Graphs and Dynamic Programming
The syntax will change, but the mindset won't.