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# Question from RHE book

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Which of the following expressions results in a positive value in x
1. int x = -1; x = x >>> 5;
2. int x = -1; x = x >>> 32;
3. byte x = -1; x = x >>> 5;
4. int x = -1; x = x >> 5;
Which is the correct answer and how? Please all java gurus experts are needed.

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1. int x = -1; x = x >>> 5;
binary representation of -1
11111111 11111111 11111111 11111111
-1>>>5
right shift replaces 1 with 0
00000111 11111111 11111111 11111111
value is large positive value

2. int x = -1; x = x >>> 32;
binary representation of -1
11111111 11111111 11111111 11111111
-1>>>32
shifts the bits 32 places hence value remains same negative
3. byte x = -1; x = x >>> 5;
results in compiler error loss of precision
4. int x = -1; x = x >> 5;
11111111 11111111 11111111 11111111
-1>>5
>> replaces 1 with 1
hence values remains large negative value
hope this helps
zarina
[ August 14, 2002: Message edited by: zarina mohammad ]
[ August 14, 2002: Message edited by: zarina mohammad ]
[ August 14, 2002: Message edited by: zarina mohammad ]

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-1>>>32 is equivalent to -1>>>0 which is -1, not 0.
In general, if i is an int, then
i >>> j
is the same as
i >>> (j % 32)
The same is true for >> and << .
If , instead, i is a long, then
i >>> j
is the same as
i >>> (j % 64)
and again, the same is true for << and >> .

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hi ron,
i edited the error. hope the answer is correct now
zarina

Greenhorn
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Excuse me! May I ask something?
Why does a byte value loose precision?
If byte has 8 bits, I can say that byte x = -1 is:
1111 1111
right?
So,if you make x >>> 5, then it should be:
0000 0111
Can you explain it to me?
Thanks!

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