21 操作符重载

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
      
posted @ 2020-09-24 13:39  nxgy  阅读(71)  评论(0编辑  收藏  举报