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From: Keith Thompson <Keith.S.Thompson+u@gmail.com>
Newsgroups: comp.lang.c
Subject: Re: does char *str="abcd"; alloc addressable memory?
Date: Mon, 06 Jul 2020 23:25:20 -0700
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John Forkosh <forkosh@panix.com> writes:
> Keith Thompson <Keith.S.Thompson+u@gmail.com> wrote:
>> <snip>
>> Some compilers have non-conforming options to treat string literals as
>> const.  (gcc has "-Wwrite-strings".)
>
> Wasn't aware of that warning, and sure enough...
>    bash-5.0$ cc -Wwrite-strings tester.c -o tester
>    tester.c: In function 'main':
>    tester.c:7:15: warning: initialization discards 'const' qualifier
>                   from pointer target type [-Wdiscarded-qualifiers]
>     7 |   char *str = ( argc>1? argv[1] : "01234&56789" );
>       |               ^
> And (as suggested downthread) changing line 7 to
>    char *str = ( argc>1? argv[1] : (char *)"01234&56789" );
> eliminates the warning, but the program still segfaults.
>
> So, the really nice -switch would be one that reverts to the
> classical behavior where "string literals" aren't const.
> Any way to do that?

I'm not aware of any such option.

If you really want a writable copy of a string literal, you might
consider strdup(), which uses malloc() to create a copy of its argument
string.  It's not C standard (though I think it's being added in the
next standard), but it is defined by POSIX -- and it's easy enough to
write your own version.

Or just create a writable array object:

    char my_string[] = "01234&56789";

I really wouldn't advise trying to modify the string literal itself,
even if you can find a compiler that lets you get away with it.

-- 
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
Working, but not speaking, for Philips Healthcare
void Void(void) { Void(); } /* The recursive call of the void */