Fun with templates

[Daniel Murphy]

"Manu" <turkeyman at gmail.com> wrote in message 
news:mailman.1754.1373081562.13711.digitalmars-d at puremagic.com...
> On 6 July 2013 11:41, Daniel Murphy <yebblies at nospamgmail.com> wrote:
>
>>
>> "Manu" <turkeyman at gmail.com> wrote in message
>> news:mailman.1752.1373074509.13711.digitalmars-d at 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).
>

Yeah it's possible I didn't finish reading your post.