In c++ 11, how to invoke an arbitrary callable object?

This is exactly what std::invoke does, but it won't be standard until C++17. You can make your own version, but it can be pretty complicated if it is fully general.

Here's the basic idea for two cases (code taken from cppreference.com):

template <class F, class... Args>
inline auto INVOKE(F&& f, Args&&... args) ->
    decltype(std::forward<F>(f)(std::forward<Args>(args)...)) {
      return std::forward<F>(f)(std::forward<Args>(args)...);
}

template <class Base, class T, class Derived>
inline auto INVOKE(T Base::*pmd, Derived&& ref) ->
    decltype(std::forward<Derived>(ref).*pmd) {
      return std::forward<Derived>(ref).*pmd;
}

Rather than implementing INVOKE yourself, use one of the library features that uses it. In particular, std::reference_wrapper works. Thus you can have the effect of std::invoke(f, args...) with std::ref(f)(args...):

template<typename F, typename... Args>
auto invoke(F f, Args&&... args)
    -> decltype(std::ref(f)(std::forward<Args>(args)...))
{
    return std::ref(f)(std::forward<Args>(args)...);
}

I didn't forward f because std::reference_wrapper requires that the object passed in is not an rvalue. Using std::bind instead of std::ref doesn't fix the problem. What this means is that for a function object like this:

struct F
{
    void operator()() && {
        std::cout << "Rvalue\n";
    }
    void operator()() const& {
        std::cout << "Lvalue\n";
    }
};

invoke(F{}) will print Lvalue, whereas std::invoke(F{}) in C++17 would print Rvalue.

_{I found the technique from this paper}