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

On 2024-08-01, Mark Summerfield <mark@qtrac.eu> 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";

>
The "this is a test" object is a literal.  It is part of the
program's image.  When you try to change it, you're making your
program self-modifying.
>
The ISO C language standard doesn't require implementations to
support self-modifying programs;  the behavior is left undefined.
>
It could work in some documented, reliable way, in a given
implementation.
>
It's the same with any other constant in the program.  [...]