C++ Function pointer wrapper throwing error when invoking - c++

Question

I've been having some trouble recently with an invoker class I've created. Reason is so I can parse an address through the class constructor and have it handle calling methods that are in memory. Problem is, when I go to return a class object type, it throws an error "Cannot convert from std::string to InvokerType". Any help is appreciated.
template <class InvokerType>
class Invoker {
public:
Invoker(std::intptr_t address) :
address(address)
{ }
template<class... Arguments>
InvokerType invoke(Arguments&&... parameters) {
InvokerType func = (InvokerType)this->address;
return (InvokerType)func(std::forward<Arguments>(parameters)...);
}
private:
std::intptr_t address;
};
std::string test(const std::string& str) {
std::cout << str << std::endl;
return "I have returned";
}
int main(void) {
using test_print_type = std::string (*)(const std::string& str);
Invoker<test_print_type> get_value{ (std::intptr_t)test };
auto ret = get_value.invoke("Hello I am a testing function\n");
std::cout << ret << std::endl; // Errors here
std::system("PAUSE");
return 0;
}

Answer 1

If you want to return the value from the invoked function, I think you need to separate the return type into its own template parameter.
#include <cstdint>
#include <string>
#include <iostream>
#include <utility>
template <class ReturnType, class...Args>
class Invoker {
public:
Invoker(std::intptr_t address) :
address(address)
{ }
ReturnType invoke(Args&&... parameters) {
auto func = (ReturnType(*)(Args...)) address;
return func(std::forward<Args>(parameters)...);
}
private:
std::intptr_t address;
};
std::string test(const std::string& str) {
std::cout << str << std::endl;
return "I have returned";
}
int main(void) {
Invoker<std::string, const std::string&> get_value{ (std::intptr_t)test };
auto ret = get_value.invoke("Hello I am a testing function\n");
std::cout << ret << std::endl; // Errors here
return 0;
}