In which circumstances Java reference equality could be different to equals() equality for an object of a type which has not overridden equals()?

Question

Is there any magic hanging around anywhere that could mean that

(object0 == object1) != (object0.equals(object1))

where object0 and object1 are both of a certain type which hasn't overridden Object.equals()?

Answer 1

No. That's exactly the definition of Object. equals() .

...this method returns true if and only if x and y refer to the same object (x == y has the value true) ...

public boolean equals( Object o ) { 
   return this == o;
}

Answer 2

Yes, if by "The type of object0 doesn't override Object.equals() " you mean the specific type and not a superclass.

If object0 and object1 are of type B, B extends A, and A overrides equals(Object obj) but B doesn't, then it is possible that B doesn't override equals(Object obj) but (object0 == object1) != (object0.equals(object1)) .

Answer 3

Well, if object0 == null and object1 == null , the first will deliver true , and the second a NullPointerException . Apart from that, there should be no observable difference.

Answer 4

Although the objects don't override equals() themselves, it is possible that one of superclasses of the object overrides the equals() method.

If you are using eclipse: open the object.java file and press control-o twice. Type ' equals ' and check if you only see one ' equals ' method: the equals method of Object.

Answer 5

The Object.java src defines its equals method as;

 return (this == obj)

so no :-)

Answer 6

是的， null == null为真，但null.equals(null) 。

Answer 7

不，如果equals()没有被覆盖，如果对象是内存中的相同对象，则返回 true。

Answer 8

No. The actual class of object0 (not necessarily the declared type of the variable) must have overridden equals() . Try printing out:

object0.getClass()

Answer 9

Here is the source code for Object.equals:

public boolean equals(Object obj) {
  151           return (this == obj);
  152       }
  153   

So, No.