在地图中存储具有不同签名的函数

Question

I am trying to create a map with string as key and a generic method as value in C++, but I do not know if that is even possible. I would like to do something like that:

void foo(int x, int y)
{
   //do something
}

void bar(std::string x, int y, int z)
{
   //do something
} 

void main()
{
   std::map<std::string, "Any Method"> map;

   map["foo"] = &foo;      //store the methods in the map
   map["bar"] = &bar;

   map["foo"](1, 2);       //call them with parameters I get at runtime
   map["bar"]("Hello", 1, 2);
}

Is that possible? If yes, how can I realise this?

Answer 1

You can type-erase the function types into a container, then provide a template operator() . This will throw std::bad_any_cast if you get it wrong.

NB because of the type erasure, you will have to specify exactly matching arguments at the call site, as eg std::function<void(std::string)> is distinct from std::function<void(const char *)> , even though both can be called with a value like "Hello" .

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

template<typename Ret>
struct AnyCallable
{
    AnyCallable() {}
    template<typename F>
    AnyCallable(F&& fun) : AnyCallable(std::function(fun)) {}
    template<typename ... Args>
    AnyCallable(std::function<Ret(Args...)> fun) : m_any(fun) {}
    template<typename ... Args>
    Ret operator()(Args&& ... args) 
    { 
        return std::invoke(std::any_cast<std::function<Ret(Args...)>>(m_any), std::forward<Args>(args)...); 
    }
    std::any m_any;
};

template<>
struct AnyCallable<void>
{
    AnyCallable() {}
    template<typename F>
    AnyCallable(F&& fun) : AnyCallable(std::function(fun)) {}
    template<typename ... Args>
    AnyCallable(std::function<void(Args...)> fun) : m_any(fun) {}
    template<typename ... Args>
    void operator()(Args&& ... args) 
    { 
        std::invoke(std::any_cast<std::function<void(Args...)>>(m_any), std::forward<Args>(args)...); 
    }
    std::any m_any;
};

void foo(int x, int y)
{
   std::cout << "foo" << x << y << std::endl;
}

void bar(std::string x, int y, int z)
{
   std::cout << "bar" << x << y << z << std::endl;
} 

using namespace std::literals;

int main()
{
   std::map<std::string, AnyCallable<void>> map;

   map["foo"] = &foo;      //store the methods in the map
   map["bar"] = &bar;

   map["foo"](1, 2);       //call them with parameters I get at runtime
   map["bar"]("Hello, std::string literal"s, 1, 2);
   //map["bar"]("Hello, const char *literal", 1, 2); // bad_any_cast
   map["bar"].operator()<std::string, int, int>("Hello, const char *literal", 1, 2); // explicit template parameters

   return 0;
}

Answer 2

The most (I cannot say best here) you can do is to use a signature erasure. That mean to convert the pointer to functions to a common signature type, and then convert them back to the correct signature before using them.

That can only be done in very special use cases (I cannot imagine a real world one) and will be highly unsecure: nothing prevent you to pass the wrong parameters to a function. In short: NEVER DO THIS IN REAL WORLD CODE .

That being said, here is a working example:

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

typedef void (*voidfunc)();

void foo(int x, int y)
{
    std::cout << "foo " << x << " " << y << std::endl;
}

void bar(std::string x, int y, int z)
{
    std::cout << "bar " << x << " " << y << " " << z << std::endl;
}

int main()
{
    std::map<std::string, voidfunc> m;
    m["foo"] = (voidfunc) &foo;
    m["bar"] = (voidfunc)& bar;
    ((void(*)(int, int)) m["foo"])(1, 2);
    ((void(*)(std::string, int, int)) m["bar"])("baz", 1, 2);
    return 0;
}

It gives as expected:

foo 1 2
bar baz 1 2

I could not find in standard whether this invokes or not Undefined Behaviour because little is said about function pointer conversions, but I am pretty sure that all common compilers accept that, because it only involve function pointers casting.

Answer 3

You cannot store functions with different signatures in a container like map , no matter if you store them as a function pointer or std ::function<WHATEVER> . The information about the signature of the function is one and only one in both cases.

The types for the value in map is one, meaning that the object stored in it are all of the same type .

So if your functions have all the same signature, then it's easy, otherwise, you have to abandon type safety and start walking in a very dangerous realm. The one in which you erase the type information about the functions stored inside the map. This translates to something like map<string, void*> .