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From: James Kuyper <jameskuyper@alumni.caltech.edu>
Newsgroups: comp.lang.c
Subject: Re: relearning C: why does an in-place change to a char* segfault?
Date: Thu, 1 Aug 2024 12:02:30 -0400
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On 8/1/24 04:06, 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";

"In translation phase 7, a byte or code of value zero is appended to
each multibyte character sequence that results from a string literal or
literals. 89) The multibyte character sequence is then used to
initialize an array of static storage duration and length just
sufficient to contain the sequence. ..." (6.4.5p6)

"... If the program attempts to modify such an array, the behavior is
undefined." (6.4.5p7).

This gives implementation the freedom,for instance, to store that array
in read-only memory, though they don't have to do so. The segfault you
got suggests that the implementation you're using did so. On other
platforms, writes to read-only memory might be silently ignored. On a
platform where it is possible to write to such memory, the
implementation is still free to optimize the code on the assumption that
you won't. That could produce bizarrely unexpected behavior if you
actually do modify it.

What you want to do is initialize an array with the static literal:

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

Nominally, such an array is initialized by copying from the string
literal's array. However, there's no way for strictly conforming code to
determine whether or not there are two such arrays. If the "text" array
has static storage duration, the string literal's array is likely to be
optimized away.