🚀 go-pugleaf

In article <Pine.ULT.3.91.951019204530.7241B-100000@mars.dsv.su.se>, Jens M Andreasen <jens-and@dsv.su.se> writes:
>static int i = 1 ? 3 : ( 1, 2);
>
>On 19 Oct 1995, Norman Diamond wrote:
>
>> No, after parsing the expression, the compiler ends up with something like:
>> 1 ? 3 : (1, 2);
>
> 1  The programmer tells that the condition is always thrue. So there is no 
> ?  condition left for the compiler to evaluate? Fine, then don't! At this
>    stage the ? operator is reduced to an elegant illusion.
> 3  This constant is for real.
> :
> (  Start of parenthesis, hmmm , interresting ...
> 1  Now, what to do with this? Consider it done and gone .
> ,  (and this cute little comma goes down the drain as well :)
> 2  This constant would have been the value of the parenthesized expression
> )  if the condition had been explicitly false.
>
>So at this stage the compiler will be left with one welldefined constant. 
>It did not even have to compute it ...



How is any of the above relevant? You've merely demonstrated that the above
expresion can be computed at compile time. You haven't demonstrated that the
expression is permitted there in the first place. Others have demonstrated
that it isn't permitted, because comma operators are not allowed in constant
expressions.

>
>> The question is not the result of evaluating the expression, the question is
>> whether the expression is permitted.  ...
>---------------------------------<snip>--------------------------------
>
>A constant is permitted, right?

Yep. Too bad we don't have one.






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