小编PoP*_*PoP的帖子

假设我们有这样的宏

#define FOO(type,name) type name

我们可以使用哪个

FOO(int, int_var);

但并不总是那么简单:

FOO(std::map<int, int>, map_var); // error: macro "FOO" passed 3 arguments, but takes just 2

我们当然可以这样做:

 typedef std::map<int, int> map_int_int_t;
 FOO(map_int_int_t, map_var); // OK

这不是非常符合人体工程学的.必须处理类型不兼容的问题.知道如何用宏来解决这个问题吗？

在OO,一个通常与接口实现回调:(粗示例)

class Message {}

class IMsgProcessor {
public:
     virtual void handle_msg(const Message& msg) = 0;
}

class RequestMsgProcessor : public IMsgProcessor {
     virtual void handle_msg(const Message& msg)  {
     // process request message
    }
}

class CustomSocket {
public:
   Socket(IMsgProcessor* p) : processor_(p) {}

   void receive_message_from_network(const Message& msg) {
       // processor_ does implement handle_msg. Otherwise a compile time error. 
       // So we've got a safe design.
       processor_->handle_msg(msg);
   }
private:
   IMsgProcessor* processor_;
}

到现在为止还挺好.使用C++ 11,另一种方法是让CustomSocket只接收std :: function对象的实例.它并不关心它的实现位置,或者即使对象是非null值:

class CustomSocket {
public:
   Socket(std::function<void(const …

考虑这个有三个构造函数的类:

class Circle {

public:
 Circle(int r) {
      _radius = r;
 }

Circle(const Circle& c){
    _radius = c.radius();
    cout << endl << "Copy constructor with lvalue reference. Radius: " << _radius;
}

Circle(Circle&& c){
    _radius = c.radius();
    cout << endl << "Copy constructor with rvalue reference. Radius:" << _radius;
}

int radius() const {
    return _radius;
}

private:
    int _radius;
};

int main() {
     Circle c1(2);
     Circle c2(c1);
     cout << endl << c2.radius(); 
     Circle c3(Circle(4));
     cout << endl << c3.radius(); …