传递字符串作为参考为一个函数指针C++

Question

下面是我的代码

#include "stdafx.h"  
#include <string.h>  
#include <iostream.h> 

using namespace std; 

class ToDoCommands 
{ 
public:  
    void getCommand(string); 
}; 

void ToDoCommands::getCommand(string command) 
{ 
    cout<<command; //here i get ping 
    void (*CommandToCall)(void); 
    CommandToCall = command; // error here i want something like 
           // CommandToCall = ping 
    CommandToCall(); 
} 

void ping(void) 
{ 
    cout<<"ping command executed"; 
} 

int main() 
{ 
    ToDoCommands obj; 
    obj.getCommand("ping"); 
} 

函数指针应参考函数平动态。与函数名称相同的字符串被传递给main中的getCommand函数。

Answer 1

C++只是不这样工作。如果你确实需要类似的东西，你必须做的功能表中按名称索引：

#include <assert.h> 
#include <iostream> 
#include <map> 
#include <string> 

using std::cout; 
using std::string; 
using std::map; 

void ping(void) 
{ 
    cout << "ping command executed\n"; 
} 

class ToDoCommands 
{ 
    public: 
    typedef void (*FunctionPtr)(); 
    typedef string Name; 

    void registerFunction(Name name,FunctionPtr); 
    void callFunction(Name); 

    private: 
    map<Name,FunctionPtr> func_map; 
}; 

void ToDoCommands::registerFunction(Name name,FunctionPtr func_ptr) 
{ 
    func_map[name] = func_ptr; 
} 

void ToDoCommands::callFunction(Name name) 
{ 
    assert(func_map.find(name)!=func_map.end()); 
    func_map[name](); 
} 

int main(int argc,char **argv) 
{ 
    ToDoCommands to_do_commands; 

    to_do_commands.registerFunction("ping",ping); 
    to_do_commands.callFunction("ping"); 
    return 0; 
} 

Answer 2

void ping(void)  
{ 
    // LL DD…DD XX 
    cout<<"ping command executed"<<endl; 
} 

class ToDoCommands  
{ 
public: 
    void getCommand(void (*CommandToCall)(void)); //getCommand(ping) 
}; 

void ToDoCommands::getCommand(void (*CommandToCall)(void)) 
{ 
    void (*CommandToCall1)(void); 
    CommandToCall1 = CommandToCall; 
    CommandToCall1(); 
} 

int main() 
{ 
    ToDoCommands obj; 
    obj.getCommand(ping); 
    return 0; 
} 

我想这和它的工作:)