assignment9 Aldin Aksay

assignment9.txt

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+Aldin Aksay
+1. Compare malloc() and calloc()
+	Calloc zero-initializes the buffer to 0, while malloc leaves the memory unitialized. Also, malloc takes in a single argument, while calloc needs two arguments. Both malloc and calloc return  pointers to allocated memory on success, or NULL on failure. They are both functions of memory allocation. 
+
+2. How would you determine the size of an allocated portion of memory?
+	You can use the sizeof operator.
+
+3. Why do you have to free memory on the heap, but not on the stack?
+	On the stack, the memory it allocated during compilation of the program, and the freeing of blocks is systematic. On the heap, the memory is allocated during the runtime.
+
+4. Why do you need to test the return value from malloc?
+	You need to test it because it might overcommit memory. Also, the elements are not intialized.

linkedlist_4.c

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+#include <math.h>
+#include <stdlib.h>
+#include <stdio.h>
+
+typedef struct node {  //defining by structure 
+	int x;
+	struct node *next;} node_t;
+
+node_t *head = NULL;   //first node is head
+node_t *current = NULL;//declares a current address
+
+void create_link(){
+	head = malloc(sizeof(node_t));		//allocates first node
+	head -> x = 5;		//first node value will be 5
+	head-> next = malloc(sizeof(node_t));  //allocating second node
+	head -> next -> x = 10;
+	head -> next -> next = malloc(sizeof(node_t));
+	head -> next -> next -> x = 15;
+}
+
+void printList(){
+	int node_num;
+	if (head == NULL)		//error situation		
+		printf("error");
+	current = head;
+	while (current != NULL){  //going through all of the nodes in the list until it reaches NULL
+	node_num++; //which node number you are on
+	printf("Node number %d has data value %d\n", node_num, current->x);
+	current = current -> next; //goes to next node
+	};
+	current = head;
+}
+
+void add(int user_x){
+	node_t *new_beg = NULL; //declares a new beginning address
+	new_beg = malloc(sizeof(node_t));  //allocating spaces for this new address
+	new_beg -> x = user_x;    
+	new_beg -> next = head; //showing that new_beg will be first in the list, before head
+	head = new_beg;  //moving head to the new beginning 
+}
+
+void remoove(int user_y){
+	current = head;  //starting at the first node
+	node_t *temp = current; //declaring a new node temp as equal to current
+	if (head->x == user_y){
+		head = head->next;
+		free(temp);
+	};
+
+	current = head;
+	temp = current;
+
+	while (current->next != NULL){
+		if (current->next->x == user_y){
+			current->next = current->next->next;
+			free (temp);
+		};
+
+		current = current->next;
+	};
+}
+
+void clear_list(){
+	current = head;
+
+	while (current->next != NULL){
+		node_t *temp = current;
+		current = current->next;
+		free (temp);
+	};
+
+}
+
+void smallest(){ 
+	int node_num;
+	int node_num2;
+	int node_sum;
+	int node_sum2;
+	node_t *temp = head;/*start of first list*/
+	while (temp != NULL){
+	node_num ++;
+	node_sum = temp -> x + node_sum;
+	temp = temp -> next;
+	};
+	printf("first list length is %d\n", node_num);
+	printf("sum of first list is %d\n", node_sum);
+
+	temp = head->next; /*start of second list*/
+	while (temp != NULL){
+	node_num2 ++;
+	node_sum2 = temp -> x + node_sum2;
+	temp = temp -> next;
+	};
+	printf("second list length is %d\n", node_num2);
+	printf("sum of second list is %d\n", node_sum2);
+	if (node_num < node_num2)
+		printf("The first list is the shortest\n");
+	if (node_num > node_num2)
+		printf("The second list is the shortest\n");
+	}
+
+
+int main(){
+create_link();
+node_t *store_head1 = head;
+printList();
+add(0);
+printList();
+remoove(5);
+printList();
+//clear_list();
+//printList();
+head = store_head1;
+smallest();
+return 0;
+}
+