In what circumstances can Java-referenced equality be different from equality equals () for an object of type that does not override equals ()?

Question

Is there any magic hanging anywhere that could mean that

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

where object0 and object1 are defined types that have not redefined Object.equals ()?

Answer 1

No. This is exactly the definition of Object. equals () .

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

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

Answer 2

Yes, if the type object0does not redefine Object.equals()"you mean a specific type, not a superclass.

object0 object1 B, B A, A equals(Object obj), B , , B equals(Object obj), (object0 == object1) != (object0.equals(object1)).

Answer 3

, object0 == null object1 == null, true, - NullPointerException;-) , .

Answer 4

equals(), , equals()...

eclipse: object.java control-o. "equals" , "equals": equals

Answer 5

Object.java src equals :

 return (this == obj)

: -)

Answer 6

Yes, null == nulltrue, but null.equals(null)not defined.

Answer 7

No, if equals()not overridden, it returns true if the objects are the same identical objects in memory.

Answer 8

No. The actual object class 0 (not necessarily a declared variable type) must have overridden equals (). Try printing 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.