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From: Bonita Montero <Bonita.Montero@gmail.com>
Newsgroups: comp.lang.c
Subject: Re: relearning C: why does an in-place change to a char* segfault?
Date: Sun, 4 Aug 2024 15:52:59 +0200
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Am 01.08.2024 um 23:07 schrieb Bart:
> 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};

This char arrays are modifyable because they're not const.

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