Question

I have the following smart_ptr class.

template <typename T>
class smart_ptr
{
public:
    // ... removed other member functions for simplicity
    T* get() { return ptr; }

    template <typename... Args>
    decltype(T::template operator ()(Args()...)) operator ()(Args... args) const
    {
        return (*get()).operator ()(args...);
    }

private:
    T* ptr;
};

However, when I use the smart_ptr class for type T without an operator (), it fails to compile (and rightly so).

What I wanted is to use std::enable_if to enable the member function only when T has operator (). It got obscure and complex very fast when I got into doing it, as I was thinking of using SFINAE to detect if T has operator () and then combine it with std::enable_if to get what I want. However, I got lost in creating SFINAE that is capable of detecting a variadic templated operator () (i.e. NOT a duplicate of Is it possible to write a template to check for a function's existence?).

Can anyone help? Or give another (perhaps simpler) solution?

p.s. It has to work on GCC 4.5.3.

EDIT: For the sake of completeness, I’ve answered my own question with a working solution for GCC 4.5.3.

Answer 1

Finally, I’ve found a workaround for GCC 4.5.3!

template <typename T>
class smart_ptr
{
public:
    // ... removed other member functions for simplicity
    T* get() const { return ptr; }

private:
    template <typename... Args>
    struct opparen_constret
    {
        typedef decltype(std::declval<T const>()(std::declval<Args>()...)) type;
    };

    template <typename... Args>
    struct opparen_ret
    {
        typedef decltype(std::declval<T>()(std::declval<Args>()...)) type;
    };

public:
    template <typename... Args>
    typename opparen_constret<Args...>::type operator ()(Args... args) const
    {
        return (*get())(args...);
    }

    template <typename... Args>
    typename opparen_ret<Args...>::type operator ()(Args... args)
    {
        return (*get())(args...);
    }

private:
    T* ptr;
};