C++ 面向对象编程：递增重载

这里有两个，分别是前置++，和后置++，具体重载可见下面代码：

#include<iostream>
using namespace std;

class test {
	friend ostream& operator<<(ostream& msg3, test msg4);
public:
	test():msg1(0),msg2(0){}
	test(int msg1, int msg2) {
		this->msg1 = msg1;
		this->msg2 = msg2;
	}
	test& operator++() { // 前置++重载
		this->msg1 += 1;
		return *this;
	}
	test operator++(int) { // 后置++重载
		test t = *this;
		this->msg1 += 1;
		return t;
	}
private:
	int msg1;
	int msg2;
};

ostream& operator<<(ostream& msg3, test msg4) {
	msg3 << msg4.msg1 << " " << msg4.msg2 ;
	return msg3;
}

int main() {
	test t(10, 10);
	cout << ++t << endl;
	cout << t << endl;
	cout << t++ << endl;
	cout << t << endl;
	return 0;
}

但是上面的写法有点繁琐，每一次调用重载的++，都需要进行拷贝构造，下面这个是加了引用的代码：

#include<iostream>
using namespace std;

class test {
	friend ostream& operator<<(ostream& msg3, const test &msg4);
public:
	test():msg1(0),msg2(0){}
	test(int msg1, int msg2) {
		this->msg1 = msg1;
		this->msg2 = msg2;
	}
	test& operator++() { // 前置++重载
		this->msg1 += 1;
		return *this;
	}
	test operator++(int) { // 后置++重载
		test t = *this;
		this->msg1 += 1;
		return t;
	}
private:
	int msg1;
	int msg2;
};

ostream& operator<<(ostream& msg3, const test &msg4) {
	msg3 << msg4.msg1 << " " << msg4.msg2 ;
	return msg3;
}

int main() {
	test t(10, 10);
	cout << ++t << endl;
	cout << t << endl;
	cout << t++ << endl;
	cout << t << endl;
	return 0;
}