C++解析(17)：操作符重载

0.目录

1.操作符重载

2.完善的复数类

3.小结

1.操作符重载

下面的复数解决方案是否可行？

示例1——原有的解决方案：

#include <stdio.h>

class Complex
{
    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); // c1 + c2

    printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB());

    return 0;
}

运行结果为：

[root@bogon Desktop]# g++ test.cpp
[root@bogon Desktop]# ./a.out
c3.a = 4, c3.b = 6

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：

示例2——使用操作符重载函数代替原有函数：

#include <stdio.h>

class Complex
{
    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 = operator + (c1, c2); // c1 + c2

    printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB());

    return 0;
}

运行结果为：

[root@bogon Desktop]# g++ test.cpp
[root@bogon Desktop]# ./a.out
c3.a = 4, c3.b = 6

可以将操作符重载函数定义为类的成员函数：

• 比全局操作符重载函数少一个参数（左操作符）
• 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

示例3——不使用友元，而是使用类的成员函数（编译器优先在成员函数中寻找操作符重载函数）：

#include <stdio.h>

class Complex
{
    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;
        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;
}

运行结果为：

[root@bogon Desktop]# g++ test.cpp
[root@bogon Desktop]# ./a.out
Complex operator + (const Complex& p)
c3.a = 4, c3.b = 6

2.完善的复数类

复数类应该具有的操作：

利用操作符重载：

• 统一复数与实数的运算方式
• 统一复数与实数的比较方式

先实现Complex.h头文件：

// Complex.h
#ifndef _COMPLEX_H_
#define _COMPLEX_H_

class Complex
{
    double a;
    double b;
public:
    Complex(double a = 0, double b = 0);
    double getA();
    double getB();
    double getModulus();

    Complex operator + (const Complex& c);
    Complex operator - (const Complex& c);
    Complex operator * (const Complex& c);
    Complex operator / (const Complex& c);

    bool operator == (const Complex& c);
    bool operator != (const Complex& c);

    Complex& operator = (const Complex& c);
};

#endif

再实现Complex.cpp具体操作：

// Complex.cpp
#include "Complex.h"
#include "math.h"

Complex::Complex(double a, double b)
{
    this->a = a;
    this->b = b;
}

double Complex::getA()
{
    return a;
}

double Complex::getB()
{
    return b;
}

double Complex::getModulus()
{
    return sqrt(a * a + b * b);
}

Complex Complex::operator + (const Complex& c)
{
    double na = a + c.a;
    double nb = b + c.b;
    Complex ret(na, nb);

    return ret;
}

Complex Complex::operator - (const Complex& c)
{
    double na = a - c.a;
    double nb = b - c.b;
    Complex ret(na, nb);

    return ret;
}

Complex Complex::operator * (const Complex& c)
{
    double na = a * c.a - b * c.b;
    double nb = a * c.b + b * c.a;
    Complex ret(na, nb);

    return ret;
}

Complex Complex::operator / (const Complex& c)
{
    double cm = c.a * c.a + c.b * c.b;
    double na = (a * c.a + b * c.b) / cm;
    double nb = (b * c.a - a * c.b) / cm;
    Complex ret(na, nb);

    return ret;
}

bool Complex::operator == (const Complex& c)
{
    return (a == c.a) && (b == c.b);
}

bool Complex::operator != (const Complex& c)
{
    return !(*this == c);
}

Complex& Complex::operator = (const Complex& c)
{
    if( this != &c )
    {
        a = c.a;
        b = c.b;
    }

    return *this;
}

最后实现main函数：

// test.cpp
#include <stdio.h>
#include "Complex.h"

int main()
{
    Complex c1(1, 2);
    Complex c2(3, 6);
    Complex c3 = c2 - c1;
    Complex c4 = c1 * c3;
    Complex c5 = c2 / c1;

    printf("c3.a = %f, c3.b = %f\n", c3.getA(), c3.getB());
    printf("c4.a = %f, c4.b = %f\n", c4.getA(), c4.getB());
    printf("c5.a = %f, c5.b = %f\n", c5.getA(), c5.getB());

    Complex c6(2, 4);

    printf("c3 == c6 : %d\n", c3 == c6);
    printf("c3 != c4 : %d\n", c3 != c4);

    (c3 = c2) = c1;

    printf("c1.a = %f, c1.b = %f\n", c1.getA(), c1.getB());
    printf("c2.a = %f, c2.b = %f\n", c2.getA(), c2.getB());
    printf("c3.a = %f, c3.b = %f\n", c3.getA(), c3.getB());

    return 0;
}

运行结果为：

[root@bogon Desktop]# g++ test.cpp Complex.cpp
[root@bogon Desktop]# ./a.out
c3.a = 2.000000, c3.b = 4.000000
c4.a = -6.000000, c4.b = 8.000000
c5.a = 3.000000, c5.b = 0.000000
c3 == c6 : 1
c3 != c4 : 1
c1.a = 1.000000, c1.b = 2.000000
c2.a = 3.000000, c2.b = 6.000000
c3.a = 1.000000, c3.b = 2.000000

注意事项：

• C++规定赋值操作符（=）只能重载为成员函数
• 操作符重载不能改变原操作符的优先级
• 操作符重载不能改变操作符的个数
• 操作符重载不应改变操作符的原有语义

3.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
• 复数的概念可以通过自定义类实现
• 复数中的运算操作可以通过操作符重载实现
• 赋值操作符只能通过成员函数实现
• 操作符重载的本质为函数定义