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

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. 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.