如何在容器中存储具有不同签名的功能对象？

Question

所以想象我们有2个功能(void : ( void ) ),(std::string : (int, std::string))而且我们还可以有10个功能.所有(或其中一些)采用不同的参数类型,并可以返回不同的类型.我们想将它们存储在a中std::map,因此我们得到这样的API:

//Having a functions like:
int hello_world(std::string name, const int & number )
{
    name += "!";
    std::cout << "Hello, " << name << std::endl;
    return number;
}
//and
void i_do_shadowed_stuff()
{
    return;
}

//We want to be capable to create a map (or some type with similar API) that would hold our functional objects. like so:
myMap.insert(std::pair<std::string, fun_object>("my_method_hello", hello_world) )
myMap.insert(std::pair<std::string, fun_object>("my_void_method", i_do_shadowed_stuff) )
//And we could call tham with params if needed:
int a = myMap["my_method_hello"]("Tim", 25);
myMap["my_void_method"];

我想知道如何将许多不同的功能放入同一个容器中.具体来说,如何使用Boost在C++ 03中执行此操作.

API应该独立于实际的函数类型(int a = myMap["my_method_hello"]("Tim", 25);没有int a = myMap<int, (std::string, int)>["my_method_hello"]("Tim", 25);).

Answer 1

#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); }
};

用法:

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 不是很漂亮.更好的建议是不要做任何事情,就像你正在尝试做的那样.

请注意,这要求具有相同参数的所有函数返回相同的类型.