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>>> operator

 
Greenhorn
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I read somewhere that >>> operator always results in a positive number.
Is it statement true or false. I think it is true but answer was given as false. Maybe the exam creator had in mind if a negative number is >>> by 0 then asnwer will be negative?
 
Greenhorn
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Hi
After the >>> operation if the result is assigned to int variable it results in positive value.But if the result needs to be casted to byte/short it may result in negative value.
correct me if i am wrong.
Regards
jaya s k
 
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hi:
You are right. After the >>> operator x times, the leading x bits is 0s(the left most bits). However, if you cast it to lower number of bits, it might result in a negative number. here is an example:
00000000 00000000 00000011 11111111 // int x = 1023
1- x >>> 2 = 255 // 00000000 00000000 00000000 11111111
2- byte y = (byte) x; // 11111111
In statement 2, we trimmed the int to a byte, which result in a negative number.
 
Greenhorn
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>>> operator may be -ve example
-1>>>32
in the above case o times we shifted it so the result is remain same as 32/32 we get 0
 
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I believe Manu is onto the point of the question. The shift operators do nothing if the amount to shift is a multiple of 32 in the case of shifting an int, or 64 in the case of a long. This creates the exception to the always-positive rule for >>>.
 
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