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From: Bart <bc@freeuk.com>
Newsgroups: comp.lang.c
Subject: Re: relearning C: why does an in-place change to a char* segfault?
Date: Thu, 1 Aug 2024 22:07:16 +0100
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On 01/08/2024 21:59, Keith Thompson wrote:
> Bart <bc@freeuk.com> writes:
>> On 01/08/2024 09:38, Richard Harnden wrote:
>>> 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.
>>
>> That's not the reason for the segfault in this case.
> 
> I'm fairly sure it is.
> 
>>                                                       With some
>> compilers, you *can* modify it, but that will permanently modify that
>> string constant. (If the code is repeated, the text is already in
>> capitals the second time around.)
>>
>> It segfaults when the string is stored in a read-only part of the binary.
> 
> A string literal creates an array object with static storage duration.
> Any attempt to modify that array object has undefined behavior.

What's the difference between such an object, and an array like one of 
these:

  static char A[100];
  static char B[100]={1};

Do these not also have static storage duration? Yet presumably these can 
be legally modified.