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

From Tim Rentsch <tr.17687@z991.linuxsc.com>
Newsgroups comp.lang.c
Subject Re: relearning C: why does an in-place change to a char* segfault?
Date 2026-03-23 21:14 -0700
Organization A noiseless patient Spider
Message-ID <86ikal50et.fsf@linuxsc.com> (permalink)
References (5 earlier) <20240801174256.890@kylheku.com> <v8i9o8$2oof8$1@dont-email.me> <v8j808$2us0r$1@dont-email.me> <864j7oszhu.fsf@linuxsc.com> <10poav2$3563c$1@raubtier-asyl.eternal-september.org>

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Bonita Montero <Bonita.Montero@gmail.com> writes:

> Am 14.08.2024 um 02:46 schrieb Tim Rentsch:
>
>> The C language doesn't have integer literals.  C has string
>> literals, and compound literals, and it has integer constants.
>> But C does not have integer literals.
>
> There's no problem when you call them integer literals.
> Except when you're pedantic as hell.

Oh no!  An insult from a narcissist!  Whatever shall I do?

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Re: relearning C: why does an in-place change to a char* segfault? Bonita Montero <Bonita.Montero@gmail.com> - 2026-03-22 09:58 +0100
  Re: relearning C: why does an in-place change to a char* segfault? Tim Rentsch <tr.17687@z991.linuxsc.com> - 2026-03-23 21:14 -0700
    Re: relearning C: why does an in-place change to a char* segfault? David Brown <david.brown@hesbynett.no> - 2026-03-24 08:58 +0100

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