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

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.