std::function with templated Argument Types

Question

struct Functor
{
public:
    template <typename... argtypes>
    Functor(std::function<void()> Function)
    {
        this->Function = Function;
    }

    ~Functor() {}

    void operator()() const
    {
        OutputDebugStringA("Hello there, General Kenobi");
        Function();
    }

private:
    std::function<void()> Function;
};

void gtk()
{
    OutputDebugStringA("What is happening to meeee");
}
Functor Draw = Functor(&gtk);
void foo() { Draw(); }

How can I make Functor's std::function accept Argument Types? I tried the following:

Functor(std::function<void(argtypes...)> Function)
Functor Draw = Functor<void>(&gtk);

But the compiler complains about 'typename not allowed'.

No, it isn't. Besides missing header includes and `OutputDebugStringA` which is OS specific, the example as posted does compile. You should post the version that reproduces the error message. If you're asking what I think you're asking, the answer is *a function pointer is not the same as `std::function`* — Praetorian, Jan 25 '18 at 20:50
No, it isn't: a complete example require a `main()` and the includes — max66, Jan 25 '18 at 20:50

score 1 · Accepted Answer · answered Jan 25 '18 at 20:58

You need to make Functor itself a template, rather than just its constructor. The parameters are part of the calling convention and are thus needed at a wider scope than just the ctor. The std::function member also needs the argument types and you also need them when actually invoking the stored callable.

std::function with templated Argument Types

1 Answers1