checking object reference equality using == (in Java) [duplicate]

Answer 1

String extends the Object type, like any other class, but does not inherit StringBuilder. Hence in your comparison String falls back to Object and they have a type in common.

Answer 2

In second example String is also an Object since all classes subclass Object and you can check if they are ==.

In first example you compare String to StringBuilder which are different classes and neither is subclass of the other so the operator fails.

Answer 3

Acording to the Java Language Specification (15.21.3):

It is a compile-time error if it is impossible to convert the type of either operand to the type of the other by a casting conversion (§5.5). The run-time values of the two operands would necessarily be unequal (ignoring the case where both values are null).

Which means, in order to "compare" two reference types, one should be able to be cast to the other. String "is" an object, but there is no "cast" between String and StringBuilder.

Answer 4

The compiler tries to help you.

When the == can never be true because of conflicting types, it assumes you made a mistake and will refuse to compile the code.

Happens with instanceof as well, and also with casts.

String a = null;
if (a instanceof StringBuilder){} // compile-error
StringBuilder b = (StringBuilder) a;  // compile-error

Answer 5

Because, String is an Object but String is not StringBuilder. You can compare reference using == on same type and supertype. Look at following example

class A{

}
class B extends A{

}

class C{

}

Now,

  A a =new A();
  A b =new B(); // or B b =new B();
  C c= new C();

  System.out.println(a==b); // It is ok
  System.out.println(a==c); // It will generate compile time error

Answer 6

Because both Strings and StringBuilders are children of the Object class. They are however not the same class, so they cannot be compared using ==. You would need to call the toString method on StringBuilder.