On 6 July 2013 11:41, Daniel Murphy <yebbl...@nospamgmail.com> wrote:
> > "Manu" <turkey...@gmail.com> wrote in message > news:mailman.1752.1373074509.13711.digitalmar...@puremagic.com... > > 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)(T _a) { Unqual!T a = _a; ... } > That doesn't do what I want at all. The signature is still f(T) not f(Unqual!T).
