c: inserting new nodes to a singly linked list using a function

Question

I used a function to insert new nodes to my singly-linked list but when i print out all the values inside the nodes after insertion, i only get the value of the first node:

// Make list
createList(head, 17);

// Insert to list
for (int x = 9; x > 0; x /= 3)
{
    if (!insertToList(head, x))
    {
        fprintf(stderr, "%s", error);
        return 1;
    }
}

The function:

bool insertToList(NODE *head, int value)
{
    NODE *node = malloc(sizeof(NODE));
    if (node == NULL)
        return false;

    node -> number = value;
    node -> next = head;
    head = node;
    return true;
}

-- Output: 17

When i don't use a function, everything works as expected:

// Make list
createList(head, 17);

// Insert to list
for (int x = 9; x > 0; x /= 3)
{
    NODE *node = malloc(sizeof(NODE));
    if (node == NULL)
    {
        fprintf(stderr, "%s", error);
        return 1;
    }

    node -> number = x;
    node -> next = head;
    head = node;
}

-- Output: 1 3 9 17

Why ?

That's because you have modified only a copy of the head pointer. — ilotXXI
– ilotXXI, Commented Oct 28, 2017 at 20:17

Parijat Purohit · Accepted Answer · 2017-10-28 20:16:33Z

1

You are passing the pointer in the function, updating it and not returning it back in which case, the outside function can never know if the head has changed. You must update the head appropriately in the for loop as well.

In the case where you don't use the function, your for loop knows the correct address ofhead every time you insert.

Probably if you return the head pointer and properly update that, it should solve your problem.

answered Oct 28, 2017 at 20:16

Parijat Purohit

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1 Comment

taoufeqreda Over a year ago

Thank you very much, your solution solved my problem.

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