C Multiple Indirection

Question

I am currently watching course about pointers in C and I have a question about multiple indirection. I understood what multiple indirection is and that is ok, but I ran on piece of code and I tried to do the same thing on a little different way and it is not working, and I wounder why? My question is about pointer casting on line 32, why this doesn't work when I cast it like: printf("Value pointed to by gp is:%s\n",(char *)gp); Here is the code:

#include <stdio.h>

int data[3];
char *words[3];

int main(int argc, char **argv)
{
        void *gp;

        char *word = "rijec";
        printf("%s\n",word);

        for(int i = 0; i < 3;i++)
        {
                data[i] = i;
        }
        words[0] = "zero";
        words[1] = "one";
        words[2] = "two";

        gp = data;
        printf("\nAddress of array data is:%p\n",gp);
        for(int i = 0; i < 3; i++)
        {
                printf("Value pointed to by gp is %d\n",*(int *)gp);
                gp = (int*)gp+1;
        }
        gp=words;
        printf("\nAddress of array of strings words is:%p\n",gp);
        for(int i = 0;i < 3; i++)
        {
                printf("Value pointed to by gp is:%s\n",*(char **)gp);
                gp  = (char **)gp+1;
        }

        return 0;
}

You can't use `(char *)gp` there, because `gp` points to the address where it expects pointers to strings, not the strings themselves. So you need to dereference the pointer first. — Cheatah, Oct 26 '21 at 18:05
ALSO: I think you probably meant `gp = (int*)gp+i;` and `gp = (char **)gp+i;`, *NOT" "gp+1". Because the constant "1" never changes in the loop :( — paulsm4, Oct 26 '21 at 18:16
ohh... the code in the text differs from the code block. Nice trick .... or something :-( — Support Ukraine, Oct 26 '21 at 18:33

martinkunev · Answer 1 · 2021-10-26T18:10:16.977

0

On line 32 gp contains the address where the array words is stored (because the array words decays to pointer during the assignment). The array words consists of 3 pointers. When you cast this to char * you tell the compiler to treat this array as if each of its elements is a char and not a pointer. You need to dereference the pointer to get the actual string.

edited Oct 26 '21 at 18:10

answered Oct 26 '21 at 18:05

martinkunev

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score 0 · Answer 2 · answered Oct 26 '21 at 18:36

When you do

gp=words;

words is converted to a pointer to the first element of the array. The first element of the array is a char*. Consequently, words is converted to a char**. So gp will hold a pointer of type char**

That's why this:

printf("Value pointed to by gp is:%s\n",(char *)gp);
                                        ^^^^^^^^
                                        wrong cast

is wrong

and this:

printf("Value pointed to by gp is:%s\n",*(char **)gp);
                                        ^ ^^^^^^^^
                                        |  correct cast
                                        |
                                        correct dereferencing

is correct

C Multiple Indirection

2 Answers2