功能图c ++ [英] Map Of functions c++

问题描述

我已经绘制了功能图.所有这些函数都是无效的，并且接收单个字符串参数.

I have made a map of functions. all these functions are void and receive single string parameter.

代码:

void f1(string params){...}
void f2(string params){...}
void f3(string params){...}

map<string , void*> funcMap;

funcMap["f1"] =(void*)&f1;
funcMap["f2"] =(void*)&f2;
funcMap["f3"] =(void*)&f3;

我该如何调用函数? 我尝试了下一个代码，但是id不起作用:

how do i call a function? I tried the next code, but id doesn't work:

void (*func)(string) =  &funcMap[commandType];
func(commandParam);

我收到此错误消息:

Server.cpp:160:46: error: cannot convert ‘void**’ to ‘void (*)(std::string) {aka void (*)(std::basic_string<char>)}’ in initialization

推荐答案

typedef void (*pfunc)(string);

map<string, pfunc> funcMap; 
funcMap["f1"] = f1; //and so forth

然后致电:

pfunc f = funcMap[commandType];
(*f)(commandParam);   

通常，为什么要扔掉类型安全性?如果它是函数指针的映射，则将其声明为一个.

In general, why throw away type safety? If it's a map of function pointers, declare it to be one.

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