如何从 2 个 void 指针调用 class 成员 function

Question

So for this function:

template<typename T>
T popObject(ScriptObject obj) { //Preforms a cast from void* on each to its type.
    assert(obj.getType() == std::type_index(typeid(T))); //Ensure type Safety.
    return obj.get<T>();
}

template <typename C, typename...Args>
    void bind(const char* name, C* context, void(C::* function)(Args...)) {
        std::vector<std::type_index> arguments = { std::type_index(typeid(Args))... };
        auto functor = [](void* func,void* contex, std::vector<ScriptObject> stack) {
            std::size_t idx = 0;
            //Call class context -> function (Args...)
        };
        _functions[name] = new BoundFunction(functor, void_cast(function), context, arguments);
    }

I'm unsure of the syntax for calling the member function.

Iv tried:

union MF{
    void(C::* pf)(Args...);
    void* p;
} mf;
mf.p = func;
contex->*(mf.pf)(popObject<Args>(stack[idx++])...);

Which does not compile, Stating: "Does not evaluate to a function taking 1 Arguments."

Tried std::invoke, But I Don't know how to Specify the Member function Signature for arg1.

So If i have a reference to the member function, And a reference to the instance of the class, As well as the template information to Derive type information. How do I call the Classes member function?

Answer 1

Thanks @Igor Tandentnik

Final Solution was:

(static_cast<C*>(ctx)->*(mf.pf))(popObject<Args>(stack[idx++])...);

As Igor pointed out, it needed to be ->* instead of.*, and added parenthesis, as -> has a lower priority then the Function call operator.

Also, thanks @TheFloatingBrain for pointing out the extra static_cast