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*.