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simple doubt - Please help !!

 
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let's say

Byte b=20;

if(b==20){
// something
}

/*
in statement b==20, b will be unboxed to byte or 20 will be boxed to Byte
*/

please explain.

Regards,

Abdul Mohsin
[ May 10, 2007: Message edited by: Abdul Mohsin ]
 
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Byte b=20;

if(b==20){
// something
}

/*
in statement b==20, b will be unboxed to byte or 20 will be boxed to Byte
*/

please explain.



Definitely primitive is not going to be boxed before comparison. b will be unboxed before comparison. It is the only way keep this rule OK. ==
comparison results true if one wrapper and another primitive comprises same value. And this comparison is done after unboxing the wrapper. This will always result true.





Thanks,
 
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I agree. Boxing 20 would always return false, as you'd compare two different Byte references.
 
Abdul Mohsin
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Thanks Chandra and Satou for your help !!!

Abdul Mohsin
 
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Originally posted by Satou kurinosuke:
I agree. Boxing 20 would always return false, as you'd compare two different Byte references.



Ah, but IF it was boxing a value in the range [-128..127] it would be caching the objects, wouldn't it? So you would get true in this case.

But it doesn't box, it unboxes. The rules are in the Java Language Specification.
 
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I think,
unbox first, otherwise you would be comparing references.

Are there any rules? I couldn't find it in JLS.

Autoboxing:


output is :
Less than Check : true
Greater than Check : true
Equality Check : false
 
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