[Issue 6714] [tdpl] Function literal does not convert to "function" and "delegate" types

d-bugmail at puremagic.com d-bugmail at puremagic.com
Sat Dec 24 09:33:02 PST 2011


http://d.puremagic.com/issues/show_bug.cgi?id=6714



--- Comment #3 from Andrei Alexandrescu <andrei at metalanguage.com> 2011-12-24 09:32:58 PST ---
(In reply to comment #2)
> (In reply to comment #0)
> > Consider:
> > 
> > 
> > void foo (int function (int, int) a){}
> > void bar (int delegate (int, int) a){}
> > 
> > void main ()
> > {
> >     foo((a, b) { return a +b;});
> >     bar((a, b) { return a +b;});
> > }
> > 
> > Neither call works. The literal does not convert to the function or the
> > delegate type.
> 
> This works though:
>      bar((int a, int b) { return a + b;});
> Is that what you meant? If it is, then this is a duplicate of 3235.
> 
> Or are argument types supposed to be deduced? If so, that's a major,
> complicated feature and difficult to implement, I think it requires an extra
> level of argument matching.

When I discussed with Walter the matter a while ago, a possible approach was
that the literal relying on deduction defines a local template function. For
example, the code:

    foo((a, b) { return a +b;});

would be lowered into:

    static auto __lambda(T1, T2)(T1 a, T2 b) { return a + b; }
    foo(__lambda);

The "static" is present because the compiler figured the lambda uses no state
from the enclosing context. Built-in conversion mechanisms should take over
from here on.

This also brings the issue of automatically converting a function to a
delegate. Walter disagrees with that, but I disagree with the basis of his
disagreement.

> It would need to be considered very carefully.
> Eg, which of these calls are ambiguous?
> 
> void bar ( double delegate(int, int) a ) {}
> void bar ( int delegate (int, int) a ){}
> 
> bar( (a, b) { return 1.0; } );

Works, goes to first overload.

> bar( (a, b) { return 1.0f; } );

Doesn't work, no conversion possible for either overload.

> bar( (a, b) { return a; } );

Depends on a's type.

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