How to store functional objects with different signatures in a container?

Question

So imagine we had 2 functions (void : ( void ) ) and (std::string : (int, std::string)) and we could have 10 more. All (or some of them) take in di

Answer 1

You can do it by casting function pointers to void pointers and back. You're expected to know the signature during run-time, so it wouldn't be an issue to hard-wire the casting operators. However the logic of doing it escapes me. It doesn't make sense at all, at least in C++. Using template classes/functions or structs of function pointers makes much more sense.

for example, with templates:

template <typename X> foo(X param1) { /* do something with param1*/};
template <typename X, typename Y> foo(X param1, Y param2)
   {/* do something with 2 params*/};
template <int X> foo(X param1) { /* only one parameter, which is int */};

now:

foo(5); // calls the third one
foo("5"); // calls the first one
foo("5", 5); // calls the second one.

Who needs a map?

Answer 2

Complete example here and here.

We can use type erasure like boost::any does. But since we want to hold only functions, we can come up with more convenient usage.

The first thing we need is function signature deduction. See here.

template <typename T>
struct function_traits : public function_traits<decltype(&T::operator())>
{};

template<typename R, typename ...Args> 
struct function_traits<std::function<R(Args...)>>
{
    typedef R result_type;
    typedef typename std::function<R(Args...)> type;
    typedef typename std::function<void(Args...)> typeNoRet;
};

template<typename R, typename ...Args> 
struct function_traits<R(*)(Args...)>
{
    typedef R result_type;
    typedef typename std::function<R(Args...)> type;
    typedef typename std::function<void(Args...)> typeNoRet;
};

template<typename R, typename cls, typename ...Args> 
struct function_traits<R(cls::*)(Args...)>
{
    typedef R result_type;
    typedef typename std::function<R(Args...)> type;
    typedef typename std::function<void(Args...)> typeNoRet;
};

To implement type erasure we need base class which is not template, and derived template class. FunctionHolder will store base class pointer. Constructor will use function_traits to determine correct derived class type.

class FunctionHolder {
    private:
        struct BaseHolder {
            BaseHolder() {}
            virtual ~BaseHolder() {}
        };

        template <typename T>
        struct Holder : public BaseHolder {
            Holder(T arg) : mFptr(arg) {}

            template<typename... Args>
            void Call(Args&&...args) {
                mFptr(std::forward<Args>(args)...);
            }

            template<typename R, typename... Args>
            R CallRet(Args&&...args) {
                return mFptr(std::forward<Args>(args)...);
            }

            T mFptr;
        };

    public:
        template<typename T>
        FunctionHolder(T t) : mBaseHolder(new Holder<typename function_traits<T>::type>(t))
                            , mBaseHolderNoRet(new Holder<typename function_traits<T>::typeNoRet>(t)) {}

        template<typename T, typename...Args>
        FunctionHolder(T&& t, Args&&... args) : mBaseHolder(new Holder<typename function_traits<T>::type>
                                                    (std::bind(std::forward<T>(t), std::forward<Args>(args)...)))
                                              , mBaseHolderNoRet(new Holder<typename function_traits<T>::typeNoRet>
                                                    (std::bind(std::forward<T>(t), std::forward<Args>(args)...))) {}

    void operator()() {
        this->operator()<>();
    }

    template<typename... Args>
    void operator()(Args&&... args) {
        auto f = dynamic_cast<Holder<std::function < void(Args...) > >*>(mBaseHolderNoRet.get());
        if (f) {
            f->Call(std::forward<Args>(args)...);
            return;
        }
        throw std::invalid_argument("");
    }

    template<typename R, typename... Args>
    R call(Args&&... args) {
        auto f = dynamic_cast<Holder<std::function<R(Args...)>>*>(mBaseHolder.get());
        if (f) {
            return f->template CallRet<R>(std::forward<Args>(args)...);
        }
        throw std::invalid_argument("");
    }

    private:
        std::unique_ptr<BaseHolder> mBaseHolder;
        std::unique_ptr<BaseHolder> mBaseHolderNoRet;
};

In this case FunctionHolder stores 2 pointers of the BaseHolder, the first one with the correct signature and second one returning void. To avoid overheads you may remove one of them, if your are going to always specify return value or never use it.

And finally we may use it like this.

struct st0 {
    st0(int x) : mX(x) {}

    std::string  print(int p) {
        std::cout << "st0::print " 
                  << mX << " " << p << std::endl;
        return "ret_from_st0::print";
    }
    int mX;
};

struct st1 {
    st1(int x) : mX(x) {}

    void operator()() {
        std::cout << "st1::operator() " 
                  << mX << " " << std::endl;
    }
    int mX;
};

void Func0(int a, int b) {
    std::cout << "Func0. "
              << " a: " << a
              << " b: " << b << std::endl;
}

void Func1(int a, int b, std::string str) {
    std::cout << "Func0. "
              << " a: " << a
              << " b: " << b
              << " str: " << str << std::endl;
}

uint64_t Func2(int a, int b, std::string str) {
    std::cout << "Func0. "
              << " a: " << a
              << " b: " << b
              << " str: " << str << std::endl;
    return 0xBAB0CAFE;
}

int main() {
    try {
        // void(int, int)
        FunctionHolder ex1(&Func0);
        ex1(1,12);
    
        // void(int, int, std::string)
        FunctionHolder ex2(&Func1);
        ex2(1, 12, std::string("Some text here"));
        
        // int(int, int, std::string)
        // call and print return value
        FunctionHolder ex3(&Func2);
        std::cout << "Ret: " << std::hex << ex3.call<uint64_t>(123, 3211, std::string("another text")) 
                  << std::dec << std::endl;
        // call and drop return value
        ex3(123, 3211, std::string("another text"));
    
        // Hold std::function<void(int)>
        std::function<void(int)> ex4 = std::bind(&Func0, 1, std::placeholders::_1);
        FunctionHolder c(std::function<void(int)>(std::bind(&Func0, 1, std::placeholders::_1)));
        ex4(12);
    
        // will bind to st0 member function print
        st0 st0object(8955);
        FunctionHolder ex5(&st0::print, st0object, std::placeholders::_1);
        ex5(2222);
        // call and print return value
        std::cout << "Ret: " << ex5.call<std::string>(7531) << std::endl;
    
        // wrap lambda function with std::function and pass to holder
        FunctionHolder ex6(std::function<void(void)>([]() {std::cout << "lambda function called" << std::endl;}));
        ex6();
    
        // functor object st1
        FunctionHolder ex7(st1(123654));
        ex7();
        
        // Will throw, because st1::operator() gets no arguments
        ex7(123);
    } catch (std::invalid_argument &e) {
        std::cout << "Invalid argument(s) were passed" << std::endl;
        // error handling here...
    }

    return 0;
}

FunctionHolder may store C like function pointers, functor objects, lambda functions and member function pointers. There are only 2 exceptions to remember.

1. If you are going to use std::bind to bind object pointer to member function, you may pass all arguments to the FunctionHolder constructor which will do binding for you. If you want to bind any other function argument(s) you have to bind it manually, construct std::function and pass it to the FunctionHolder.

FunctionHolder fh(std::function<void(std::string)>(std::bind(&Func1, 888, 333, std::placeholders::_1)));
fh(std::string("Ok."));

2. To store lambda functions, first wrap them with the std::function, as in example above.

In case of passing wrong arguments, FunctionHolder will throw std::invalid_argument.

Answer 3

The thing is, somehow, when you call your functions, you already know what type they will be.

If we do something like

int x = map["key"](1, "2")

we can already deduce that whatever function is stored in "key" is of type (int (*)(int, char*)) so we might as well have done something like

int x = map_of_int_and_string_to_int["key"](1, "2");

and avoid all the hassle of merging all the keys together... While it is true that C++ has some overloading features precisely for this kind of stuff I can't really see why you should bother in this particular case.

And in the end, why would you want to put all those functions in the same map in the first place? They don't share any similar interfaces so you can't uniformly access them, you can't iterate over them and you can't opaquely pass them around to someone else. Without anything in common there is nothing you can safely do to the functions in this hypothetical map.

Answer 4

#include <functional>
#include <iostream>
#include <string>
#include <map>

class api {
    // maps containing the different function pointers
    typedef void(*voidfuncptr)();
    typedef int(*stringcrintptr)(std::string, const int&);

    std::map<std::string, voidfuncptr> voida;
    std::map<std::string, stringcrintptr> stringcrint;
public:
    // api temp class
    // given an api and a name, it converts to a function pointer
    // depending on parameters used
    class apitemp {
        const std::string n;
        const api* p;
    public:
        apitemp(const std::string& name, const api* parent)
            : n(name), p(parent) {}
        operator voidfuncptr()
        { return p->voida.find(n)->second; }
        operator stringcrintptr()
        { return p->stringcrint.find(n)->second; }
    };

    // insertion of new functions into appropriate maps
    void insert(const std::string& name, voidfuncptr ptr)
    { voida[name]=ptr; }
    void insert(const std::string& name, stringcrintptr ptr)
    { stringcrint[name]=ptr; }
    // operator[] for the name gets halfway to the right function
    apitemp operator[](std::string n) const
    { return apitemp(n, this); }
};

Usage:

api myMap; 

int hello_world(std::string name, const int & number )
{
    name += "!";
    std::cout << "Hello, " << name << std::endl;
    return number;
}

int main()
{
    myMap.insert("my_method_hello", &hello_world );
    int a = myMap["my_method_hello"]("Tim", 25);
}

http://ideone.com/SXAPu Not very pretty. Better advice is to not do anything even remotely like whawtever it is you're trying to do.

Note that this requires all functions with the same parameters to return the same type.

Answer 5

You can use boost::any...

#include <boost/any.hpp>
#include <iostream>
#include <map>
#include <string>

void voidFunc()
{
    std::cout << "void called" << std::endl;
}

void stringFunc(std::string str)
{
    std::cout << str << std::endl;
}

int main()
{
    std::map<std::string, boost::any> funcs;
    funcs.insert(std::pair<std::string, boost::any>("voidFunc", &voidFunc));
    funcs.insert(std::pair<std::string, boost::any>("stringFunc", &stringFunc));

    boost::any_cast<void(*)(void)>(funcs["voidFunc"])();
    boost::any_cast<void(*)(std::string)>(funcs["stringFunc"])("hello");
    return 0;
}

Note that you will get a runtime exception if you don't specify the function signature correctly in the any_cast.