Wrappers

[Kamila Jolan]

Hello,

I would like to ask why the following code works:

   Short s1= 1;
 if(s1 < new Integer(7+1))
... 

If we use the "==" or "!=" instead of "<" or ">" the code won't compile because two different wrapper type cannot be compared. Are the operators like "<" or ">" preform an automatic cast like compound assignments(e.g. +=)??


Kindest regards,

Kamila

[swaraj gupta]

This is because in case of "==" and "!=" the primitive value is boxed to its corresponding wrapper class object to perform the reference comparison and then the comparison is performed, which is not allowed between two non-comparable object types.

[Kamila Jolan]

Swaraj thanks for the answer, I know that "==" and "!=" preform automatic boxing, but I am not sure what "<" and ">" doing with the wrappers?

[Henry Wong]

Kamila Jolan wrote: I know that "==" and "!=" preform automatic boxing, but I am not sure what "<" and ">" doing with the wrappers?

Actually, the "==" and "!=" comparison operators don't unbox the operands -- when used to compare two objects. These comparison operators will work for both primatives and objects, so when you use on two objects, it will do reference comparisons. On the other hand, if you compare one primative with an object, it knows that you want to do a value compare and will unbox the wrapper object.

In the case of the ">" and "<" comparison operators, it only works for primatives. There is no such a thing as one object being greater than another. So, it know that you want to do a value compare and will unbox both wrapper operands.

Henry

[Abimaran Kugathasan]

Henry Wong wrote:
In the case of the ">" and "<" comparison operators, it only works for primatives. There is no such a thing as one object being greater than another. So, it know that you want to do a value compare and will unbox both wrapper operands.

And, in the above mentioned case, the both unboxed values are promoted to int and then compared.

[Kamila Jolan]

Thanks a lot for help! now I got it