#include <iostream>
using namespace std;
class Number {
public:
int value;
explicit Number(int n) : value(n) { }
};
Number operator+(const Number &n1, const Number &n2) {
Number result(0);
result.value = n1.value + n2.value;
return result;
}
Number operator+(const Number &n1, int n2) {
Number result(0);
result.value = n1.value + n2;
return result;
}
int main() {
Number num1 = Number(10) + Number(20);
Number num2 = Number(10) + 5;
cout << "num1 = " << num1.value << endl;
cout << "num2 = " << num2.value << endl;
return 0;
}
函数中的参数顺序和调用顺序必须一致,如下面的调用就报错了:
我们定义Number和int相加的函数时,Number参数是在前面的,则在使用时也必须保持Number在前面。
定义的顺序也可以修改,必须让int在前面,则调用的时候int就要写在前面如下:
与加号运算符一样,可以声明为类的成员函数,也可以声明为全局函数,这里就只写一种,示例如下:
class Number {
public:
int value;
explicit Number(int n) : value(n) { }
};
Number& operator<<(Number & n1, Number & n2) {
n1.value = n1.value + n2.value;
return n1;
}
int main() {
Number n1 = Number(1);
Number n2 = Number(2);
Number n3 = Number(3);
Number result = n1 << n2 << n3;
cout << result.value << endl;
return 0;
}
注:左移运算符的返回值必须是引用类型才能连续调用左移符号,否则只能调用一次,如下:
Number operator<<(Number & n1, Number & n2) {
n1.value = n1.value + n2.value;
return n1;
}
int main() {
Number n1 = Number(1);
Number n2 = Number(2);
Number result = n1 << n2;
cout << result.value << endl;
return 0;
}
如上代码,<<函数的返回值不是引用类型的,则<<不能连续使用,如下是错误的:
另外需要注意,对象不能使用匿名对象,匿名对象无法使用<<来操作,如下是错误的:
另外,<<函数也是可以重载的,如下:
Number & operator<<(Number & n1, Number & n2) {
n1.value = n1.value + n2.value;
return n1;
}
Number & operator<<(Number & n1, int n2) {
n1.value = n1.value + n2;
return n1;
}
int main() {
Number n1 = Number(1);
Number n2 = Number(2);
Number n3 = Number(3);
Number result = n1 << n2 << n3 << 4 << 5;
cout << result.value << endl;
return 0;
}
另外,如果访问了类的私有成员,也可以设置为友元,如下:
class Number {
friend Number & operator<<(Number & n1, Number & n2);
friend Number & operator<<(Number & n1, int n2);
friend int main();
private:
int value;
public:
explicit Number(int n) : value(n) { }
};
Number & operator<<(Number & n1, Number & n2) {
n1.value = n1.value + n2.value;
return n1;
}
Number & operator<<(Number & n1, int n2) {
n1.value = n1.value + n2;
return n1;
}
int main() {
Number n1 = Number(1);
Number n2 = Number(2);
Number n3 = Number(3);
Number result = n1 << n2 << n3 << 4 << 5;
cout << result.value << endl;
return 0;
}
通过<<实现打印对象,如下:
如上代码,cout不知道如何打印Number对象,所以IDE直接就报错了,说白了就是系统库并没有提供接收Number参数的<<函数,所以我们自己可以提供一个,如下:
class Number {
public:
int value;
explicit Number(int n) : value(n) { }
};
ostream & operator<<(ostream & out, const Number & number) {
printf("number.value = %d ", number.value);
return out;
}
int main() {
Number n1(1);
Number n2(3);
cout << n1 << n2 << "hello" << endl;
return 0;
}
运行结果如下:
number.value = 1 number.value = 3 hello
class Number {
public:
int value;
explicit Number(int n) : value(n) { }
// 前++
Number & operator++() {
value++;
return *this;
}
// 后++
Number operator++(int) {
Number copy = *this; // 利用拷贝构造函数创建出一个复本
value++;
return copy;
}
};
ostream & operator<<(ostream & out, const Number & number) {
out << number.value;
return out;
}
int main() {
Number n1(1);
Number n2(1);
cout << ++n1 << endl;
cout << n2++ << endl; // 这里打印的并不是n2,而是++操作返回的一个副本对象
cout << n2 << endl;
return 0;
}
c++编译器至少给一个类添加4个函数:
示例如下:
class Number;
ostream & operator<<(ostream & out, const Number & number);
class Number {
public:
int * valuePointer;
explicit Number(int n) {
valuePointer = new int(n);
}
~Number() {
delete valuePointer;
}
Number & operator=(const Number& number) {
// 预防浅拷贝问题,编译器默认是浅拷贝,如:valuePointer = number.valuePointer
cout << "自身value = " << *valuePointer << ", 传入value = " << *number.valuePointer << endl;
*valuePointer = *number.valuePointer;
return *this;
}
};
ostream & operator<<(ostream & out, const Number & number) {
out << *number.valuePointer;
return out;
}
int main() {
Number n1(1);
Number n2(2);
Number n3(3);
n1 = n2 = n3;
cout << n1 << endl;
cout << n2 << endl;
cout << n3 << endl;
return 0;
}
运行结果如下:
自身value = 2, 传入value = 3
自身value = 1, 传入value = 3
3
3
3
这说明赋值操作的优先级是从右到左的。
class Number {
public:
int value;
explicit Number(int n) : value(n) { }
};
bool operator==(Number & n1, Number & n2) {
return n1.value == n2.value;
}
ostream & operator<<(ostream & out, const Number & number) {
out << number.value;
return out;
}
int main() {
Number n1(1);
Number n2(2);
Number n3(2);
bool result1 = n1 == n2;
bool result2 = n2 == n3;
cout << result1 << endl;
cout << result2 << endl;
return 0;
}
还有其它的关系,比如:<、>、!=、>=、<=等,这些的实现也是一样的,就不写了。
class Number {
public:
int value;
explicit Number(int n) : value(n) { }
void operator()() {
cout << "函数调用: " << value << endl;
}
};
ostream & operator<<(ostream & out, const Number & number) {
out << number.value;
return out;
}
int main() {
Number n1(1);
Number n2(2);
Number n3(3);
n1();
n2();
n3();
return 0;
}
这个函数调用很奇怪,不知道为什么要搞这种形式。还可加入参数,示例如下:
class Number {
public:
int value;
explicit Number(int n) : value(n) { }
void operator()() {
cout << "函数调用: " << value << endl;
}
void operator()(int a, string b) {
cout << "函数调用: " << value + a << b << endl;
}
};
ostream & operator<<(ostream & out, const Number & number) {
out << number.value;
return out;
}
int main() {
Number n1(1);
Number n2(2);
Number n3(3);
n1(1, "一");
n2(2, "二");
n3(3, "三");
return 0;
}