Bugs in template constraints

[Pelle]

On 08/03/2010 11:07 PM, Andrej Mitrovic wrote:
>
>
> On Tue, Aug 3, 2010 at 10:23 PM, Philippe Sigaud
> <philippe.sigaud at gmail.com <mailto:philippe.sigaud at gmail.com>> wrote:
>
>
>
>     On Tue, Aug 3, 2010 at 22:04, Andrej Mitrovic
>     <andrej.mitrovich at gmail.com <mailto:andrej.mitrovich at gmail.com>> wrote:
>
>         Oh and there's a shorter way to write this example, by using
>         isInputRange from std.range, like so:
>
>         if (isInputRange!R && is(typeof({x = fun(x, range.front);})))
>
>
>     Does this work, without the () after the } ?
>
>
> I haven't even noticed those.
>
> In the following, If I add the pair of ()'s I get void as a return type.
> If I remove them, I get void delegate():
>
> writeln(typeid(typeof( delegate void () {int x = 1;}())));    // writes void
> writeln(typeid(typeof( delegate void () {int x = 1;})));      // writes
> void delegate()
>
> So I definitely need to add them. When not added the expression
> evaluates to void delegate(), which is a valid type and the constraint
> then passes.
>
> If I understood everything, this code in the constraint:
>
> is(typeof({x = fun(x, range.front);}() )))
>
> creates an anonymous function, the compiler sees it's trying to access x
> so it makes it a delegate, and it infers that the function takes no
> arguments and the return type is void. Did I get this right?
>

You only need to call it if you want to check the return type. You 
cannot create a function with content that can't compile, so in this 
case, the () isn't needed.

Correct me if I'm wrong, of course. :)