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
