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1 需要解决的问题

• 复数运算

  class Complex
  {
  public:
      int a;
      int b;
  };
  
  int main()
  {
      Complex c1 = {1,2};
      Complex c2 = {3,4};
      Compelx c3 = c1 + c2;  //error: no match for ‘operator+’ (operand types are ‘Complex’ and ‘Complex’)
      
      return 0;
  }
  

• 示例：复数的加法操作

  • Demo

    #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());  //c3.a = 4, c3.b = 6
        
        return 0;
    }
    

2 操作符重载

• C++ 中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为系统预定义的操作符，如+，-，*，/等
  

• 示例：操作符重载

  • Demo

    #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;
        }
        
        //全局操作符重载函数
        friend Complex operator + (const Complex& p1, const Complex& p2);
    };
    
    //全局操作符重载函数
    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; // 等价于函数调用形式：operator + (c1, c2)
        
        printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB());  //c3.a = 4, c3.b = 6
        
        return 0;
    }
    

3 成员函数重载操作符

• 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 Type
  {
  public:
      Type operator Sign(const Tyep& p)
      {
          Type ret;
          return ret;
      }
  };
  

• 示例：成员函数重载操作符

  • Demo

    #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;
        }
        
        //成员函数操作符重载
        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; //等价于函数调用形式： c1.operator + (c2)
        
        printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB());
        
        return 0;
    }
    

  • 编译运行

    Complex operator + (const Complex& p)
    c3.a = 4, c3.b = 6
    

21 操作符重载

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原文地址：https://www.cnblogs.com/bky-hbq/p/13723668.html