如何创建map< string，class :: method>在c + +并能够搜索函数并调用它? [英] How to create map<string, class::method> in c++ and be able to search for function and call it?

问题描述

我正在尝试在C ++中创建字符串和方法的映射，但我不知道该怎么做.我想做类似的事情(伪代码):

I'm trying to create a map of string and method in C++, but I don't know how to do it. I would like to do something like that (pseudocode):

map<string, method> mapping =
{
  "sin", Math::sinFunc,
  "cos", Math::cosFunc,
  ...
};

...

string &function;
handler = mapping.find(function);
int result;

if (handler != NULL)
  result = (int) handler(20);

说实话，我不知道C ++是否有可能.我想要一个字符串，方法的映射，并能够在映射中搜索函数.如果存在给定的函数字符串名称，那么我想使用给定的参数来调用它.

To be honest I don't know is it possible in C++. I would like to have a map of string, method and be able to search for function in my mapping. If given string name of function exists then I would like to call it with given param.

推荐答案

好吧，我不是这里流行的Boost Lovers Club的成员，所以就这样-在原始C ++中.

Well, I'm not a member of the popular here Boost Lovers Club, so here it goes - in raw C++.

#include <map>
#include <string>

struct Math
{
    double sinFunc(double x) { return 0.33; };
    double cosFunc(double x) { return 0.66; };
};

typedef double (Math::*math_method_t)(double);
typedef std::map<std::string, math_method_t> math_func_map_t;

int main()
{

    math_func_map_t mapping;
    mapping["sin"] = &Math::sinFunc;
    mapping["cos"] = &Math::cosFunc;

    std::string function = std::string("sin");
    math_func_map_t::iterator x = mapping.find(function);
    int result = 0;

    if (x != mapping.end()) {
        Math m;
        result = (m.*(x->second))(20);
    }
}

很明显，如果我正确理解您想要的是方法指针，而不是函数/静态方法指针.

That's obviously if I have understood correctly that you want a method pointer, not a function/static method pointer.

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