Why doesn't std::reference_wrapper implicitly cast to a reference when calling member function? [duplicate]

Answer 1

This is a language limitation as @MikeSeymour pointed out.

And this is the reason why I think std::reference_wrapper should have an overloaded operator& that returns the address of the wrapped object:

template<class T>
struct reference_wrapper{
   ...
   T* operator &(){return ptr;}
}

So that later one can use &Foo->f() or (*&Foo).f() instead of Foo.get().f() or static_cast<T&>(Foo).f().

Answer 2

Because . is always used to access members of the object it's applied to. Type conversions aren't considered.

There's a proposal to allow overloading of operator., to enable exactly what you want, but that won't be standard until at least C++17, if at all.