Fun with templates

[Timon Gehr]

On 07/06/2013 03:34 AM, Manu wrote:
> Okay, so I feel like this should be possible, but I can't make it work...
> I want to use template deduction to deduce the argument type, but I want
> the function arg to be Unqual!T of the deduced type, rather than the
> verbatim type of the argument given.
>
> I've tried: void f(T : Unqual!U, U)(T a) {}
> and: void f(T)(Unqual!T a) {}
>
> Ie, if called with:
>    const int x;
>    f(x);
> Then f() should be generated void f(int) rather than void f(const int).
>
> I don't want a million permutations of the template function for each
> combination of const/immutabe/shared/etc, which especially blows out
> when the function has 2 or more args.
>
> Note: T may only be a primitive type. Obviously const(int*) can never be
> passed to int*.

void f(T)(const(T) a) {}

This will strip off const, immutable and inout, but not shared.

I'd have thought that

void f(T)(const(shared(T)) a) {}

would strip shared as well, but DMD does not match int to 
const(shared(int)) at all which I think is incorrect.