c++ - 函数指针 + 映射 + 迭代器 + 类

Question

我有个小问题!

---

class AKSH
{
private:
 typedef map<string,void (AKSH::*)()> t_list;
public:
 t_list list;
 AKSH(){...}

};

AKSH sh;

void AKSH::doWork()
{
 map<string,void (AKSH::*)()>::iterator it;
 ...
 if(it != list.end())
 {
 (sh.*it->second)();
}

---

int main()
{
 AKSH aksh;
 aksh.doWork();
}

我必须生成 2 个对象(aksh、sh)。我如何将 aksh 与迭代器一起使用，或者行 (sh.*it->second) 中是否有错误？

Answer 1

你想要:(this->*(it->second))();

但是，在添加之前，请阅读 this page !如果您不阅读该页面，并且继续使用指向成员函数的指针，我保证您会气急败坏。

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

class AKSH
{
private:
 typedef std::map<std::string,void (AKSH::*)()> t_list;
 const std::string name;
public:
 t_list list;
 void add() { std::cout << name << ": " << __FUNCTION__ << "\n"; }
 void sub() { std::cout << name << ": " << __FUNCTION__ << "\n"; }
 void doWork(const std::string&);
 AKSH(const std::string& name) :name(name) {
  list["add"] = &AKSH::add;
  list["sub"] = &AKSH::sub;
 }
};

AKSH sh("sh");

void AKSH::doWork(const std::string& str)
{
 AKSH::t_list::iterator it;
 it = list.find(str);
 if(it != list.end())
 {
  (this->*(it->second))();
 }
 else
 {
  std::cout << name << ": No such command: " << str << "\n";
 }
}
int main()
{
 AKSH aksh("aksh");
 aksh.doWork("add"); aksh.doWork("sub");
 sh.doWork("add"); sh.doWork("sub");
 sh.doWork("div"); aksh.doWork("mul");
}