How to get a pointer type's pointed type?

Question

I have a pointer type Ptr. It might be T*, unique_ptr, shared_ptr, or others. How to get its pointed type at compilation time? I try the following but failed

template<class Ptr>
void f()
{
    typedef decltype(*Ptr()) T; // give unexpected results
}

The following deleted answer works very well.

typedef typename std::remove_reference<decltype(*std::declval<Ptr>())>::type T;

Answer 1

This is one way to do it.

Create a helper class, with appropriate specializations to deduce the pointer type.

template <typename T> Pointer;

template <typename T> Pointer<T*>
{
   typedef T Type;
};

template <typename T> Pointer<shared_ptr<T>>
{
   typedef T Type;
};

template <typename T> Pointer<unique_ptr<T>>
{
   typedef T Type;
};

template<class Ptr>
void f()
{
    typedef typename Pointer<Ptr>::Type PointerType;
}

Answer 2

I was in the middle of writing up this answer when I saw someone else had already posted it, so I gave up. But then that other answer disappeared, so here it is. If the original reappears, I'll delete this.

If Ptr is (for example) int*, then decltype(*Ptr()) evaluates to int& rather than int, which is probably the cause of your error (whatever it is). Try:

std::remove_reference<decltype(*Ptr())>::type

Or, if it's possible Ptr might not have a default constructor:

std::remove_reference<decltype(*std::declval<Ptr>())>::type