Function to insert a node in a linked list in C

Question

I'm trying to learn data structures in C and I'm stuck at the first function I made. If I run this nothing happens.
I get no errors but the program doesn't print anything.


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

typedef struct node{
    int data;
    struct node *next;
}node;

void insert(int data, node *head){
    node *new_node = malloc(sizeof(node));
    new_node->data = data;
    head = new_node;
}

int main(){
    node *head = NULL;
    insert(8, head);
    printf("head.data: %d\n", head->data);
}

But if I put the code from the function insert in the main function it works.

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

typedef struct node{
    int data;
    struct node *next;
}node;

int main(){

    node *head = NULL;
    node *new_node = malloc(sizeof(node));
    new_node->data = 5;
    head = new_node;
    printf("head.data: %d\n", head->data);
}

Do I not know how to use functions in C or what is the problem with my first code?

head is passed by value to the insert function, which means that any assignment to the local head variable in that function does not affect the other head variable in main. — trincot
– trincot, Commented Mar 24, 2022 at 19:12

Vlad from Moscow · Accepted Answer · 2022-03-24 19:27:37Z

In this call

insert(8, head);

the pointer head is passed to the function by value.

It means that the function deals with a copy of the value of the pointer.

Changing the copy within the function does not reflect on the original value of the pointer.

Either you need to pass the pointer by reference through a pointer to it or to return the new value of the pointer from the function and to assign it to the original pointer.

Also you forgot to set the data member next of the created node to NULL or more precisely to head.

Here you are.

int insert( node **head, int data )
{
    node *new_node = malloc( sizeof( node ) );
    int success = new_node != NULL;

    if ( success )
    {
        new_node->data = data;
        new_node->next = *head;
        *head = new_node;
    }

    return success;
}

and the function is called like

insert( &head, 8 );

or

if ( !insert( &head, 8 ) )
{
    puts( "Error: not enough memory." );
}

Another approach is to return the new value of the pointer head from the function.

For example

node * insert( node *head, int data )
{
    node *new_node = malloc( sizeof( node ) );
    
    if ( new_node != NULL )
    {
        new_node->data = data;
        new_node->next = head;
    }

    return new_node;
}

In this case you need to use the function with caution.

For example

node *tmp = insert( head, 8 );
if ( tmp != NULL ) 
{
    head = tmp;
}
else
{
    puts( "Error: not enough memory." );
}

trincot · Accepted Answer · 2022-03-24 19:16:55Z

1

The variable head in main and the variable head in insert are two different variables. When in insert you assign something to that local variable, it does not affect the variable head in main.

You can solve this in different ways. One is to pass the address of head to the function, so insert can actually modify what is at that address:

Not your question, but you should also initialise the next member of new_node:

void insert(int data, node **headPtr){
    node *new_node = malloc(sizeof(node));
    new_node->data = data;
    new_node->next = *headPtr;
    *headPtr = new_node;
}

int main(){
    node *head = NULL;
    insert(8, &head);
    printf("head.data: %d\n", head->data);
}

answered Mar 24, 2022 at 19:16

trincot

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