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From: Tim Rentsch <tr.17687@z991.linuxsc.com>
Newsgroups: comp.lang.c
Subject: Re: does char *str="abcd"; alloc addressable memory?
Date: Tue, 07 Jul 2020 06:26:35 -0700
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John Forkosh <forkosh@panix.com> writes:

> Malcolm McLean <malcolm.arthur.mclean@gmail.com> wrote:
>
>> On Monday, 6 July 2020 12:57:06 UTC+1, Bonita Montero  wrote:
>>
>>> The compiler should give you an error for casting a string-literal to
>>> a pointer to non-const characters.  If cou cast it explicitly you may
>>> modify the caracters and that usually works, but it's not guaranteeed.
>>> Most platforms will make a writable copy of a read-only mapped page.
>>
>> It doesn't warn because const was a late addition to C, and string literals
>> were originally char *s.
>> You can declare a writeable string with
>>
>> char writeable[] = "Overwrite me";
>>
>> this is also seldom a good idea.
>
> As mentioned upthread
>    char *str = ( argc>1? argv[1] :  (char *)"01234&56789" );
> compiles without the -Wwrite-strings warning, but still
> segfaults.  And
>    char str[] = ( argc>1? argv[1] : "01234&56789" );
> gives an error
>    tester.c:7:16: error: invalid initializer
>        7 |   char str[] = ( argc>1? argv[1] : "01234&56789" );
>          |                ^
>
> As suggested downthread, I can easily
>    char default[99] = "whatever",
>         *str = ( argc>1? argv[1] : default );
> but I'd prefer the more compact notation if possible.
> Any way to get that to work?

It's compound literals to the rescue!

    char *str =   argc > 1  ? argv[1]  : (char[]){ "01234&56789" };

No errors, no undefined behavior, no problem.