Re: Conversion of delegate with default arguments to delegate with less arguments?

[Jacob Carlborg]

On 2011-09-02 21:11, Andrej Mitrovic wrote:
Code:
void main()
{
     auto foo = (int x = 10){ /* */ };
     void delegate() bar = foo;
}

Since foo takes an argument that already has a default it can be used
as a simple `void delegate()`, so maybe it makes sense for `foo` to be
able to implicitly convert to such a type.

Unfortunately doing a cast doesn't work properly:

import std.stdio;

void main()
{
     auto foo = (int x = 10){ writeln(x); };
     void delegate() bar;
     bar = cast(typeof(bar))foo;

     bar();  // prints garbage
}

So maybe this type of conversion is impossible in the first place due
to how arguments are passed? I don't know all the technical tidbits,
but from a user's point of view an implicit conversion kind of makes
sense (if it's possible).

I think it would be usable. I also think that a delegate/function that 
returns a value could be implicitly converted to a delegate/function of 
the same signature but returns void instead. I would be the same as 
calling the delegate that returns a value and not assign the returned 
value to a variable.

--
/Jacob Carlborg