一元運算子過載


下表顯示了一元運算子,其目的列表。

函式名稱 運算子 目的
opUnary - 負數(數位補)
opUnary + 相同值(或者,副本)
opUnary ~ 按位元取反
opUnary * 存取指向
opUnary ++ 自增
opUnary -- 自減

一個例子如下,解釋如何過載一個二元運算子。

import std.stdio;

class Box
{
   public:

      double getVolume()
      {
         return length * breadth * height;
      }
      void setLength( double len )
      {
         length = len;
      }

      void setBreadth( double bre )
      {
         breadth = bre;
      }

      void setHeight( double hei )
      {
         height = hei;
      }
      Box opUnary(string op)()
      {
          if(op == "++")
          {
             Box box = new Box();
             box.length = this.length + 1;
             box.breadth = this.breadth + 1 ;
             box.height = this.height + 1;
             return box;
          }
      }
   private:
      double length;      // Length of a box
      double breadth;     // Breadth of a box
      double height;      // Height of a box
};
// Main function for the program
void main( )
{
   Box Box1 = new Box();    // Declare Box1 of type Box
   Box Box2 = new Box();    // Declare Box2 of type Box
   double volume = 0.0;     // Store the volume of a box here

   // box 1 specification
   Box1.setLength(6.0);
   Box1.setBreadth(7.0);
   Box1.setHeight(5.0);

    // volume of box 1
   volume = Box1.getVolume();
   writeln("Volume of Box1 : ", volume);

   // Add two object as follows:
   Box2 = ++Box1;

   // volume of box2
   volume = Box2.getVolume();
   writeln("Volume of Box2 : ", volume);

}

讓我們編譯和執行上面的程式,這將產生以下結果:

Volume of Box1 : 210
Volume of Box2 : 336