Re: calling exit in response to an error shouldn't be an error...

[Zan Lynx <zlynx at acm dot org>]

I am not a GCC developer, just a regular user of C. But I have some 
comments below:

On 6/9/2019 3:21 PM, L A Walsh wrote:
> If I have a function returning NULL on error (including EOF).
>
> So the program calls exit if the function doesn't return
> a non-zero value (func() || exit(1)).
>
> I have:
>
> --/tmp/ex.c--
>
> main() { char * buf[512];
>      while (gets(buf) || exit(0)) puts(buf);
> }
>
> -- compile w/:
> gcc -fpermissive --no-warnings  -o /tmp/ex /tmp/ex.c
>
> Got:
> /tmp/ex.c: In function ‘main’:
> /tmp/ex.c:2:22: error: void value not ignored as it ought to be
>    while (gets(buf) || exit(0) ) puts(buf);
>                        ^~~~~~~
>
> I understand that the while is testing the value of exit
> "when it returns", but since it doesn't, why flag an error
> (if exit is part of C-standard?) but *especially*, why
> not a warning, like a type mismatch or such?i

There is a slight misunderstanding of operators here. This is not while 
testing the value of exit(0). This is the "||" operator. It needs a 
value on *both* sides of || so it can evaluate them both and return a 
boolean result. Your code has no value on the right-hand side so it's 
meaningless.

> But this:
>
> --/tmp/ex2.c--
>
> main(){ char * buf[512];
>    while (1) {
>      fgets(buf) || exit(1);
>      fputs(buf);
>    }
> }
>
> ---compile + output:
> > gcc -fpermissive --no-warnings  -o /tmp/ex2 /tmp/ex2.c
> /tmp/ex2.c: In function ‘main’:
> /tmp/ex2.c:4:18: error: void value not ignored as it ought to be
>      fgets(buf) || exit(1);
>                    ^~~~~~~
>
> Ultra confusing -- how is exit value used?  Isn't it thrown away?

Same problem here.

I haven't tested it but you might be able to make this work with a comma 
operator like "while (gets(buf) || (exit(0), 1)) puts(buf);"

Oh, and "gets" is no longer a function. It's not just deprecated. It's 
*gone*.

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