Re: relearning C: why does an in-place change to a char* segfault?

On 01/08/2024 21:59, Keith Thompson wrote:

Bart <bc@freeuk.com> writes:

On 01/08/2024 09:38, Richard Harnden wrote:

On 01/08/2024 09:06, Mark Summerfield wrote:

This program segfaults at the commented line:
>
#include <ctype.h>
#include <stdio.h>
>
void uppercase_ascii(char *s) {
      while (*s) {
          *s = toupper(*s); // SEGFAULT
          s++;
      }
}
>
int main() {
      char* text = "this is a test";
      printf("before [%s]\n", text);
      uppercase_ascii(text);
      printf("after  [%s]\n", text);
}

text is pointing to "this is a test" - and that is stored in the
program binary and that's why can't modify it.

>
That's not the reason for the segfault in this case.

 I'm fairly sure it is.
 

                                                      With some
compilers, you *can* modify it, but that will permanently modify that
string constant. (If the code is repeated, the text is already in
capitals the second time around.)
>
It segfaults when the string is stored in a read-only part of the binary.

 A string literal creates an array object with static storage duration.
Any attempt to modify that array object has undefined behavior.

What's the difference between such an object, and an array like one of these:
  static char A[100];
  static char B[100]={1};
Do these not also have static storage duration? Yet presumably these can be legally modified.