Object Solutions - D Programming Language
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# Object Solutions

#### `Object` Dersi Problem Çözümleri

1. For the equality comparison, `rhs` being non-`null` and the members being equal would be sufficient:

       enum Color { blue, green, red }

       class Point {
           int x;
           int y;
           Color color;

       // ...

           override bool opEquals(Object o) const {
               const rhs = cast(const Point)o;

               return rhs && (x == rhs.x) && (y == rhs.y);
           }
       }

2. When the type of the right-hand side object is also `Point`, they are compared according to the values of the `x` members first and then according to the values of the `y` members:

       class Point {
           int x;
           int y;
           Color color;

       // ...

           override int opCmp(Object o) const {
               const rhs = cast(const Point)o;
               enforce(rhs);

               return (x != rhs.x
                       ? x - rhs.x
                       : y - rhs.y);
           }
       }

3. Note that it is not possible to cast to type `const TriangularArea` inside `opCmp` below. When `rhs` is `const TriangularArea`, then its member `rhs.points` would be `const` as well. Since the parameter of `opCmp` is non-`const`, it would not be possible to pass `rhs.points[i]` to `point.opCmp`.

       class TriangularArea {
           Point[3] points;

           this(Point one, Point two, Point three) {
               points = [ one, two, three ];
           }

           override bool opEquals(Object o) const {
               const rhs = cast(const TriangularArea)o;
               return rhs && (points == rhs.points);
           }

           override int opCmp(Object o) const {
               auto rhs = cast(TriangularArea)o;
               enforce(rhs);

               foreach (i, point; points) {
                   immutable comparison = point.opCmp(rhs.points[i]);

                   if (comparison != 0) {
                       /* The sort order has already been
                        * determined. Simply return the result. */
                       return comparison;
                   }
               }

               /* The objects are considered equal because all of
                * their points have been equal. */
               return 0;
           }

           override size_t toHash() const {
               /* Since the 'points' member is an array, we can take
                * advantage of the existing toHash algorithm for
                * array types. */
               return typeid(points).getHash(&points);
           }
       }

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