this code gives head->next = NULL although it should not

Question

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

// linked-list implementation of stack

typedef struct stack node;

struct stack{

    int value;
    node* next;    

};

void push(node*,int);
void pop(node*);



int main(){

node* head=malloc(sizeof(node));
head->next = NULL;
push(head,5);
if (head->next == NULL){
printf("head->next is NULL.");
}
pop(head);

}


void push(node* head,int value){

    head->value = value;
    node* temp = malloc(sizeof(node));
    temp->next = head;
    head = temp;

}

The above code prints head->next = NULL although it should not because when push is called then temp->next = head and then head = temp so now the head->next should be equal to the previous head's location.

Answer 1

in see that in your push() function you want to change the head of the list but actually you are passing it by-value so it will not be affected.

node *head means that you have a pointer to a NODE and that you can actually modify only the properties of that NODE , not the head pointer itself.

you should use node **head which is a pointer to a pointer so it means that with it you can actually modify the head pointer.

That was a very short explanation... please try to read more about Pass-By-Value in C and Pass-By-Reference .

After this modification you must also pass the address of the head pointer as a parameter (&head) and make changes to it as *head ( in your function)

Here is the code:

void push(node**,int); void pop(node*);

int main(){

node* head=malloc(sizeof(node));
head->next = NULL;
push(&head,5);
if (head->next == NULL){
printf("head->next is NULL.");
}
//pop(head);

}


void push(node** head,int value){

    (*head)->value = value;
    node* temp = malloc(sizeof(node));
    temp->next = *head;
    (*head) = temp;

}