jazzabeanie - 1 year ago 40

Java Question

I am struggling to understand this Koan:

`@Koan`

public void equalsMethodCanBeChangedBySubclassesToTestsIfTwoObjectsAreEqual() {

Object object = new Integer(1);

assertEquals(object.equals(object), true);

assertEquals(object.equals(new Integer(1)), __);

// Note: This means that for the class 'Object' there is no difference between 'equal' and 'same'

// but for the class 'Integer' there is difference - see below

}

As far as I understand, because

`object`

`Object`

`.equals()`

If

`new Integer(1)`

`object`

`false`

`true`

Edit: I understand that integers between -128 and 127 are cached. If my understanding of the

`object`

Answer

Integer overrides `equals`

and checks if the underlying `int`

is equal to the `int`

of the other `Integer`

instance, and if so, returns `true`

. The reason why Integer's `equals`

method is invoked, and not the one from Object, is that the runtime type of `object`

is Integer.

Integer is an Object, but due to the overridden `equals`

, no object identity is used.

All following boolean expressions evaluate to true:

```
print((new Integer(1).equals(1)));
print((new Integer(1).equals(new Integer(1))));
print((((Integer) 1).equals(new Integer(1))));
print(((Integer) 1).equals(1));
```

Now consider autoboxing, which reuses instances for values in the range [-128,127]. The following statements about object equality are all evaluating to `true`

:

```
1 == ((Integer) 1)
((Integer) (-128)) == ((Integer) (-128)) // in autoboxing range
((Integer) (+127)) == ((Integer) (+127)) // same
((Integer) (-200)) != ((Integer) (-200)) // not autoboxing
((Integer) (+200)) != ((Integer) (+200)) // same
((Integer) (-128)) != (new Integer(-128)) // explicit new instance, so no autoboxing
((Integer) (+127)) != (new Integer(+127)) // same
```