Question

I’m sure this has been asked before, but a cursory google and stack overflow search didn’t turn up the answer.

#include <stdio.h>

int main() {
    char a[128][1024];
    strcpy(a[0], "hello");
    strcpy(a[1], "foo");
    strcpy(a[2], "bar");
    char **b = a;
    printf("%s\n", a[0]); //same as printf("%s\n", a)
    printf("%s\n", a[2]+1); //print from 2nd char of 3rd string
    printf("%s\n", b); //same as printf("%s\n", a), makes sense
    printf("%s\n", b[0]); //segfault???
}

First off, why is the last one a segfault? I’d expect same behavior as array a. How would I access the n-th string from b in a generalized way? What are the differences in treatment between a and b?

On a similar note, the way I understand it, a[n] is syntactic sugar for *(a+n). Is this correct, both for pointers and for arrays? Yet it seems getting different behavior for a and b.

Thanks!

Answer 1

char **b says “At the place where b points, b[0], there is a pointer to a char. And, if I use b[1], b[2],…, those are also pointers to char.”

In contrast char a[128][1024] says “a is 128 arrays of 1024 char.” When you do this, at the place where a is, there are no pointers. There are just char. In memory, it looks like 131,072 char in a row (128•1024 = 131,072).

When you assign char **b = a, assuming the compiler allows you to, you set b to be the address of a. When you use b[0], there should be a pointer there. But there is not. There are just char there. When you pass b[0] to printf, the compiler goes to where b points, loads several bytes as if they were a pointer, and passes the resulting value to printf. Then printf crashes, because the bytes point to some bad location.

A proper definition of b would be char (*b)[1024] = a;.