Question

This is in reference to this question: Why is a pointer to pointer needed to allocate memory in this function?

The answer to the question explained why this didn’t work:

void three(int * p)
{
    p = (int *) malloc(sizeof(int));
    *p = 3;
}

void main()
{
    int *p = 0;
    three(p);
    printf("%d", *p);
}

… but this works:

void three(int ** p)
{
    *p = (int *) malloc(sizeof(int));
    **p = 3;
}

void main()
{
    int *p = 0;
    three(&p);
    printf("%d", *p);
}

This also works, by returning a pointer from the function. Why is that?

int* three(int * p)
{
    p = (int *) malloc(sizeof(int));
    *p = 3;
    return p;
}

void main()
{
    int *p = 0;
    p = three(p);
    printf("%d", *p);
}

Answer 1

int* three(int * p) 
    {
        p = (int *) malloc(sizeof(int));
        *p=3;
        return p;
    }

Because here you’re returning a copy of the pointer p and this pointer now points to valid memory, which contains the value 3.

You originally passed in a copy of your p as an argument, so you’re not changing the one you passed in, but a copy. Then you return that copy, and assign it.

From the comment, which is a very valid point, this will also work just as well:

 int* three() 
        {
           //no need to pass anything in. Just return it.
            int * p = (int *) malloc(sizeof(int));
            *p=3;
            return p;
        }