Template deduction based on return type?

Question

I'd like to be able to use template deduction to achieve the following:

GCPtr<A> ptr1 = GC::Allocate();
GCPtr<B> ptr2 = GC::Allocate();

instead of (what I currently have):

GCPtr<A> ptr1 = GC::Allocate<A>();
GCPtr<B> ptr2 = GC::Allocate<B>();

My current Allocate function looks like this:

class GC
{
public:
    template <typename T>
    static GCPtr<T> Allocate();
};

Would this be possible to knock off the extra < A> and < B>?

Thanks

Answer 1

That cannot be done. The return type does not take part in type deduction, it is rather a result of having already matched the appropriate template signature. You can, nevertheless, hide it from most uses as:

// helper
template <typename T>
void Allocate( GCPtr<T>& p ) {
   p = GC::Allocate<T>();
}

int main()
{
   GCPtr<A> p = 0;
   Allocate(p);
}

Whether that syntax is actually any better or worse than the initial GCPtr<A> p = GC::Allocate<A>() is another question.

P.S. c++11 will allow you to skip one of the type declarations:

auto p = GC::Allocate<A>();   // p is of type GCPtr<A>

Answer 2

You could try to use a macro for it. Other than that, I don't see how that's supposed to work with just one statement.

#define ALLOC(ptrname,type) GCPtr<type> ptrname = GC::Allocate<type>()

ALLOC(ptr1,A);

Johannes' points are valid. The >> issue is easily fixed. But I think having commas as part of the type requires the C99 preprocessor varargs extension:

#define ALLOC(ptrname,...) GCPtr< __VA_ARGS__ > ptrname = GC::Allocate< __VA_ARGS__ >()

ALLOC(ptr1,SomeTemplate<int,short>);

Answer 3

In the same way you can't overload functions on return type, you can't do template deduction on it. And for the same reason - if f() is a template/overload that returns something, what type to use here:

f();

Answer 4

The only thing I can think of: make Allocate a non-template that returns a non-template proxy object that has a templated conversion operator which does the real work:

template <class T>
struct GCPtr
{

};

class Allocator
{
public:
    template <class T>
    operator GCPtr<T>() { return GCPtr<T>(); }
};

class GC
{
public:
    static Allocator Allocate() { return Allocator(); }//could give a call-back pointer?
};

int main()
{
    GCPtr<int> p = GC::Allocate();
}

Answer 5

You could go the opposite route.

If you're using an up to date compiler (MSVC 2010 which should be out in a couple of days, or the current version of GCC) and don't mind relying on C++0x features:

auto ptr1 = GC::Allocate<A>();
auto ptr2 = GC::Allocate<B>();

would save you the extra <A> and <B>, just not on the right hand side. :)