i'm trying to find the address of a function from a std::function.
The first solution was:
size_t getAddress(std::function<void (void)> function) {
    typedef void (fnType)(void);
    fnType ** fnPointer = function.target<fnType *>();
    return (size_t) *fnPointer;
}
But that only works for function with (void ()) signature, since i need for function that signature are (void (Type &)), i tried to do
template<typename T>
size_t getAddress(std::function<void (T &)> function) {
    typedef void (fnType)(T &);
    fnType ** fnPointer = function.target<fnType *>();
    return (size_t) *fnPointer;
}
And i get "Error - expected '(' for function-style cast or type construction"
Update: Is any way to capture member class address? for class members i'm using:
template<typename Clazz, typename Return, typename ...Arguments>
size_t getMemberAddress(std::function<Return (Clazz::*)(Arguments...)> & executor) {
    typedef Return (Clazz::*fnType)(Arguments...);
    fnType ** fnPointer = executor.template target<fnType *>();
    if (fnPointer != nullptr) {
        return (size_t) * fnPointer;
    }
    return 0;
}
Update: To capture lambda i'm using
template <typename Function>
struct function_traits
    : public function_traits<decltype(&Function::operator())> {
};
template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) const> {
    typedef ReturnType (*pointer)(Args...);
    typedef std::function<ReturnType(Args...)> function;
};
template <typename Function>
typename function_traits<Function>::function
to_function (Function & lambda) {
    return static_cast<typename function_traits<Function>::function>(lambda);
}
template <typename Lambda>
size_t getAddress(Lambda lambda) {
    auto function = new decltype(to_function(lambda))(to_function(lambda));
    void * func = static_cast<void *>(function);
    return (size_t)func;
}
std::cout << getAddress([] { std::cout << "Hello" << std::endl;}) << std::endl;
 
     
    