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  • 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
        ", 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
        ", 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)
        ");
                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)
        ");
            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
        ", c3.getA(), c3.getB());
            
            return 0;
        }
        
      • 编译运行

        Complex operator + (const Complex& p)
        c3.a = 4, c3.b = 6
        
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  • 原文地址:https://www.cnblogs.com/bky-hbq/p/13723668.html
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