c - pointer of a pointer in linked list append

Question

I normally program in python. To increase performance of my simulations, I am learning C. I have a problem to understand the use of a pointer of a pointer when implementing the append function to a linked list. This is an excerpt of the code from my book (Understanding Pointers in C by Kanetkar).

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

struct node{
    int data;
    struct node *link;
};

int main(){
    struct node *p; //pointer to node structure
    p = NULL;   //linked list is empty

    append( &p,1);
    return 0;
}

append( struct node **q, int num){
    struct node *temp, *r;  //two pointers to struct node
    temp = *q;

    if(*q == NULL){
        temp = malloc(sizeof(struct node));
        temp -> data = num;
        temp -> link = NULL;
        *q = temp;
    }
    else{
        temp = *q;
        while( temp -> link != NULL)
            temp = temp -> link;
        r = malloc(sizeof(struct node));
        r -> data = num;
        r -> link = NULL;
        temp -> link = r;
    }
}

In this code, I pass the double pointer **q to the append function. I get that this is the adress of the address, i.e. the adress of NULL in this case.

I just don't get why one does it like this. Would it not be valid to remove one * operator from everything in the append() function and simply pass the adress of NULL (i.e. p instead of &p) to the append() function?

I have googled this question. The answers are either too hard to understand (since I'm just a C beginner) or too plain. I'm thankful for any hints, comments or links where I can read up about this.

Answer 1

When you pass things to functions in C, whether its variables or pointers, it's a copy of the original.

Quick example:

#include <stdio.h>
void change(char *in)
{
    // in here is just a copy of the original pointer.
    // In other words: It's a pointer pointing to "A" in our main case 
    in = "B";
    // We made our local copy point to something else, but did _not_ change what the original pointer points to.
}
void really_change(char **in)
{
    // We get a pointer-to-a-pointer copy. This one can give us the address to the original pointer.
    // We now know where the original pointer is, we can make _that one_ point to something else.
    *in = "B";
}
int main(int argc, char *argv[])
{
    char *a = "A";
    change(a);
    printf("%s
", a); /* Will print A */
    really_change(&a);
    printf("%s
", a); /* Will print B */
    return 0;
}

So the first function call to change() gets passed a copy of a pointer to an address. When we do in = "B" we only change the copy of the pointer we got passed.

In the second function call, the really_change(), we get passed a copy of a pointer-to-a-pointer. This pointer contains the address to our original pointer and voila, we can now reference the original pointer and change what the original pointer should point to.

Hopefully it explains it somewhat more :)