So currently i have an if
statement in the GetNth
function that im trying to test. but when i inserted a printf
function, it made me notice that it goes through the if
statement even if the condition is not met, however, when i remove the printf
statement the program works flawlessly. any explanation would be appreciated.
Notice! This is not my code, im trying to study linked lists and was changing the code around trying to learn!
The Code:
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Takes head pointer of the linked list and index
as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
struct node* current = head;
int count = 0; /* the index of the node we're currently
looking at */
int a;
while (current != NULL)
{
if (count == index)
return(current->data);
a = current->data;
printf("\n Testing If in linked list, should bring same desired value which is 4 %d \n ",a);
count++;
current = current->next;
}
/* if we get to this line, the caller was asking
for a non-existent element so we assert fail */
assert(0);
}
/* Drier program to test above function*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
if (head != NULL)
{
}
/* Check the count function */
printf("Element at index 3 is %d", GetNth(head, 3));
getchar();
}
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