Confusion about wrapper classes.

[Vijay Tidake]

Hi All,

public class Test{
	public static void main(String[] args) {
		Double d=new Double(1);
		Double e=new Double(1);
		
		System.out.println(d+" "+e);//This line prints 1.0 1.0
		
		if(d==e){//so why this is not true?
			System.out.println("d==e");
		}
		if(d.equals(e)){
			System.out.println("d.equals(e)");
		}
	}
	
}

The programs prints the output as d.equals(e) and why not d==e,if the line above (line 6)print the values of d and e as 1.0?

[Akilan Paul]

hi
while camparing "d==e " , here you are comparing two different objects(address of two object, always different d and e) so it returns false.

d.equals(e) this returns true because you are comparing value of the objects. so it returns true.

hope you are clear now.

[Vijay Tidake]

But ,Why line 6 is giving me the values as 1.0 1.0

If Im printing the objects it should give me something like Double@hashcode, right.


And if you print the hashcode for both the 'd' and 'e'.They are same

[Dragos Nica]

vijay tidake wrote:But ,Why line 6 is giving me the values as 1.0 1.0

If Im printing the objects it should give me something like Double@hashcode, right.


And if you print the hashcode for both the 'd' and 'e'.They are same

Maybe because Double provide a default implementaion of method toString() that return the value...

[Matthew Brown]

When you print an object it's converting to a String, using the toString() method. The default version of that in the Object class does what you describe, but lots of classes override it to do something more useful. Double is one of those - it overrides it to produce the same as writing out the primitive value.

[Vijay Tidake]

So you mean to say if im camparing d==e ,Im actually doing String==String.

This is what you want to say?

[Matthew Brown]

No. If you are comparing d == e, you are checking if they a referring to the same object.

If you write:

System.out.println(d + " " + e);

This is the same as:

System.out.println(d.toString() + " " + e.toString());

[Arhaan Singhania]

Hi,

The '==' implementation you are using is inherited from the Object class as you have not ovveriden
the method. The Object class is implemented in a way which tell, whether two objects are referring to
same memory location or not. In you case, you have created two different objects, so they will never
true as their memory address are not same.

Hope this helps.
Arhaan

[Matthew Brown]

Arhaan Singhania wrote:The '==' implementation you are using is inherited from the Object class as you have not ovveriden
the method.

That's slightly misleading. The == operator is not inherited from Object, and cannot be overridden. It always checks if the two references refer to the same object.

The equals() method is inherited from Object, and in Object is defined to use ==, which is where the confusion has arisen, I think. The equals() method can be overridden.

[jishnu dasgupta]

Matthew Brown wrote:

Arhaan Singhania wrote:The '==' implementation you are using is inherited from the Object class as you have not ovveriden
the method.

That's slightly misleading. The == operator is not inherited from Object, and cannot be overridden. It always checks if the two references refer to the same object.

The equals() method is inherited from Object, and in Object is defined to use ==, which is where the confusion has arisen, I think. The equals() method can be overridden.

Exactly!!! what d and e are storing(bitwise) is not the value 1 but the reference where two Double objects are stored.It does not reflect anything about the contents of the objects.
And as Matthew has rightly pointed, when you display the two objects, the toSring() method is called, which has been inherited and overriden by the Double class, to display its contents.
so although your print statement prints 1.0 and 1.0,
the if(d==e) fails.

And again, the equals() method is similarly inherited and overriden by the Double class to check whether the content of two Double objects being compared is the same!!! so the if() condition returns true!!!

so if you had something like this

    Double d=new Double(1);
    Double e=d;
    if(d==e)   //would return true as no new object is being created and reference is being copied!!!
    {
          //do something
     }

Hope it clears your doubt!!!