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From: Richard Harnden <richard.nospam@gmail.invalid>
Newsgroups: comp.lang.c
Subject: Re: relearning C: why does an in-place change to a char* segfault?
Date: Thu, 1 Aug 2024 09:38:13 +0100
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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.

Change it to:

char text[] = "this is a test";

You can modify that, text gets it's own copy.