Passing a linked list head through a function as address in C

Question

I have a question regarding passing the head of a linked list in C through a function. So the code goes something like this:

#include <stdio.h>
//Defining a structure of the node
struct node { 
    int data;
    struct node* next;
    };

void insert (struct node* rec, int x) {
    struct node* temp = (struct node*)malloc(sizeof(struct node));
    temp->data = x;
    temp->next = NULL;
    rec = temp; // head and rec is now pointing to the same node
}

void print(struct node* rec){
    printf("%d", rec->data); //error occurs here
    puts("");
}

main(){
    struct node *head = NULL; //head is currently pointing to NULL
    insert (head, 5); //Passing the head pointer and integer 5 to insert()
    print(head);
}

So as you see, the error occurs when I tried printing rec->data. Why did the error occur? I thought since the pointer rec and head are all pointing to the same node in the heap, there should not be any problem?

Thank you.

Solution 1

You could pass a struct node** as suggested by @sje397.

However, I would suggest the following design (which, in my opinion is easier to reason about too):

/* returns the new head of the list */
struct node *insert (struct node* current_head, int x) {
    struct node* temp = (struct node*)malloc(sizeof(struct node));
    temp->data = x;
    temp->next = current_head;
    return temp;
}

and use it like

head = insert(head, 5);

In this case I would also rename the function something like push_front.

Just for completeness, I think @sje397 meant something like the following (Typical linked list code rewritten again and again by every C programmer...):

void insert(struct node **head, int x) {
    struct node* new_head = (struct node*)malloc(sizeof(struct node));
    new_head->data = x;
    new_head->next = *head;

    *head = new_head;
}

Solution 2

In C there is no pass by reference.
Your insert function isn't inserting a node in the list, its just changing the node which the head points to. Because of temp->next = NULL the list will always contain two nodes.

Another error is that you're just modifying a local copy of the head node. To fix this You have 3 choices:

-You can make the head node global

-You can pass a pointer to the head node(pointer to pointer) to the function.

-You can return the modified head node by the function.

Comments

[sje397]

That fixes the mentioned bug, but it's not an insert function.

[Alex Reynolds]

Calling this function gives you nodes that point to NULL. So if you think about how a linked list is structured, this doesn't really offer insertion functionality. You need to modify rec so that the new node points to it, or vice versa.

[Luna]

I see... @sje397 nimrodm Thank you so much for the help! This pointer to pointer thing is getting me confused every time.