Integer a=100, b=100;
System.out.printIn(a == b);
Integer c=200, d=200;
System.out.printIn(c == d);
.
.
.
The use second command print false, why?
Let's see how good you are to detect small error!!!

Where are the JAVA daemons?

Integer a=200, b=200;
System.out.println(a == b);

this will give you false as well.

int a=200, b=200;
System.out.println(a == b);

this will give you true.

Integer a=100, b=100;
System.out.println(a.equals(b));

Integer c=200, d=200;
System.out.println(c.equals(d));

this will return true in all cases

int a=100, b=100;
System.out.println(a==b);

int c=200, d=200;
System.out.println(c==d);

this will return true in all cases

Reason: you cannot compare 2 integers the same way you will compare int (they are 2 different things entirely). I ain't got time to teach now o jare.

dhtml18:

this will give you false as well.


this will give you true.




this will return true in all cases



this will return true in all cases

Reason: you cannot compare 2 integers the same way you will compare int (they are 2 different things entirely). I ain't got time to teach now o jare.

To expand upon this, when you declare Integer a = 100; you are creating an instance of Object Integer. To check for object content equality, the .equals() method should suffice.

The reason why Integer a = 100, b = 100;

and (a == b) comes out to false is because they are not pointing or referencing to the same object.

Just passing by with my 'java bible' text book firmly held in my armpit.

crotonite:
Just passing by with my 'java bible' text book firmly held in my armpit.

lol