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

Bart <bc@freeuk.com> writes:

On 01/08/2024 20:39, Kaz Kylheku wrote:

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.

>
So is the text here:
>
  char text[]="this is a test";
>
But this can be changed without making the program self-modifying.

Incorrect.  The string literal results in the creation of an array
object.  Any attempt to modify that array object would have undefined
behavior -- but there's no way to modify it because its address isn't
available to the code.

`text` is a distinct object.  At execution time (assuming it's defined
at block scope), that object is initialized by copying from the string
literal object.  (This is what happens in the abstract machine; there
are opportunities for optimization that might result in the string
literal object not existing in the generated code.)

I guess it depends on what is classed as the program's 'image'.

Not really.

Given:

    int n = 42;

you can't modify 42, but you can modify n.  There's no need to consider
the idea of self-modifying code.  You're just trying to make it seem
more confusing than it really is.

[...]

--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
void Void(void) { Void(); } /* The recursive call of the void */