需要解决的问题下面的复数解决方案是否可行？class Complex{public: int a; int b;};==>int main(){ Complex c1 = {1, 2}; Complex c2 = {3, 4}; Complex c3 = c1 + c2; return 0;}编程实验： 复数的加法操作#include <stdio.h>class Complex{private: int a; int b;public: Complex(int a = 0, int b = 0) { this->a = a; this->b = b; } int getA() { return a; } int getB() { return b; } friend Complex Add(const Complex& p1, const Complex& p2);};Complex Add(const Complex& p1, const Complex& p2){ Complex ret; ret.a = p1.a + p2.a; ret.b = p1.b + p2.b; return ret;}int main(){ Complex c1(1, 2); Complex c2(3, 4); Complex c3 = Add(c1, c2); printf(“c3.a = %d, c3.b = %d\n”, c3.getA(), c3.getB()); return 0;}输出：c3.a = 4, c3.b = 6思考：Add 函数可以解决 Complex 对象相加的问题， 但是 Complex 是现实世界中确实存在的复数，并且复数在数学运算中的地位和普通的实数相同。为什么不能让 + 操作符也支持复数相加呢？操作符重载C++ 中的重载能够扩展操作符的功能操作符的重载以函数的方式进行本质：用特殊形式的函数扩展操作符的功能通过 operator 关键字可以定义特殊的函数operator 的本质是通过函数重载操作符语法：Type operator Sign(const Type& p1, const Type& p2){ Type ret; return ret;}Sign 为系统中预定义的操作符，如：+，-，*，/ 等编程实验： 操作符重载初探#include <stdio.h>class Complex{private: int a; int b;public: Complex(int a = 0, int b = 0) { this->a = a; this->b = b; } int getA() { return a; } int getB() { return b; } friend Complex operator + (const Complex& p1, const Complex& p2);};Complex operator + (const Complex& p1, const Complex& p2){ Complex ret; ret.a = p1.a + p2.a; ret.b = p1.b + p2.b; return ret;}int main(){ Complex c1(1, 2); Complex c2(3, 4); Complex c3 = c1 + c2; // ==> operator + (c1, c2); printf(“c3.a = %d, c3.b = %d\n”, c3.getA(), c3.getB()); return 0;}可以将操作符重载函数定义为类的成员函数比全局操作符重载少一个参数不需要依赖友元就可以完成操作符重载编译器优先在成员函数中寻找操作符重载函数class Name{public: Name operator Sign(const Name& p) { Name ret; return ret; }};隐藏的 this 指针充当左操作数编程实验： 成员函数重载操作符#include <stdio.h>class Complex{private: int a; int b;public: Complex(int a = 0, int b = 0) { this->a = a; this->b = b; } int getA() { return a; } int getB() { return b; } Complex operator + (const Complex& p) { Complex ret; printf(“Complex operator + (const Complex& p)\n”); ret.a = this->a + p.a; // this 指针充当左操作数 ret.b = this->b + p.b; return ret; } friend Complex operator + (const Complex& p1, const Complex& p2);};Complex operator + (const Complex& p1, const Complex& p2){ Complex ret; printf(“Complex operator + (const Complex& p1, const Complex& p2)\n”); ret.a = p1.a + p2.a; ret.b = p1.b + p2.b; return ret;}int main(){ Complex c1(1, 2); Complex c2(3, 4); Complex c3 = c1 + c2; // ==> c1.operator + (c2); printf(“c3.a = %d, c3.b = %d\n”, c3.getA(), c3.getB()); return 0;}输出:Complex operator + (const Complex& p)c3.a = 4, c3.b = 6小结操作符重载是 C++ 的强大特性之一操作符重载的本质是通过函数扩展操作符的功能operator 关键字是实现操作符重载的关键操作符重载遵循相同的函数重载规则全局函数和成员函数都可以实现对操作符的重载以上内容参考狄泰软件学院系列课程，请大家保护原创！