Template deduction for function based on its 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>?

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>