gcc: why is "abcdef"[3] not a constant (error: initializer element is not constant)

[Ángel González]

On 08/05/15 17:19, Martin wrote:
> On 05/08/2015 08:34 AM, m-h-l wrote:
>> Funny thing is that g++, ArmC and ICC do not have a problem with 
>> "abcdef"[3]
>> as an initializer.
>> So I wonder what shall be the sense to forbid this in gcc.
>> This way I cannot use gcc to calculate e.g. CRCs or Hashs over real 
>> constant
>> things at build time without extra tools, even not in such a clear 
>> case as
>> here. With the other compilers its no problem.
>
> It is true that the C standard requires a constant expression in this
> context and that the bracket expression is not such an expression, so
> the program isn't strictly conforming.
>
> However, the C standard leaves room for implementations to provide
> extensions as long as they don't change the behavior of strictly
> conforming programs. Unless such a program can be constructed that
> would detect such an extension, providing it wouldn't make
> an implementation wrong (non-conforming). GCC has a rich set of
> extensions so there is a precedent for such a request if one should
> be made. I would encourage you to open a Bugzilla and include your
> arguments in it.
>
> Martin
+1

I'm afraid Andrew email was misguided, appearing as rude. gcc has many 
extensions, and there's no reason it couldn't add this one, which I 
agree would be useful.

I think the difference you mention is that the code is being rejected 
before doing constant folding (or simply, not accepting a string 
literal). It is funny because I had thought I had successfully done 
something like that in the past  (maybe an old version of gcc allowed 
it?) nut browsing my code, what I found was quite the opposite, a case 
where I ending up passing to a macro several character constants instead 
of a constant string precisely to work around this problem.


As for g++ accepting that syntax, I would't take for granted that it is 
doing it at build time. I think the compiler allows it just as it would 
accept:
> typedef struct
> {
>     const char  oneChar;
>     int ptr2;
> } xxx_t;
> static xxx_t ms =  { "12345"[3], *rand()* };

So I think it is accepting it because it intends to later run the code 
for its initialization (it should later notice it is actually constant 
and be able to optimize it).

If you try that construct a switch, where the compiler absolutely 
requires it to be available at build time, g++ also fails to compile it:
>   switch (argc) {
>         case "12345"[3]:
>             break;
>         default:
>             return 1;
>     }

produces
> error: an array reference cannot appear in a constant-expression


Best regards