Fun with templates

[Dmitry Olshansky]

06-Jul-2013 05:34, Manu пишет:
> 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) {}

The thing is that if even if you somehow force your way past IFTI what 
would be generated is:

f!(const int)(int arg);
f!(immutable int)(int arg);
f!(shared int)(int arg);
f!(const shared int)(int arg);

Which IMHO falls short of desired goal.
Short of using a forwarding thunk (that you don't like, but if there was 
force_inline?) we'd have to hack the compiler.

>
> 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*.




-- 
Dmitry Olshansky