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Right Shift >> 33 on an int

 
Rod Nichols
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I got this from the rules round up. The question goes like this
int c = 270;
c>>33;
What is the result ? The answer is 270>>1 but I am not sure how that is done.
270 in binary is :
0000 0000 0000 0000 0000 0001 0000 1110
When you right shift >>, the sign bit, which is the high order bit, is used to fill in the bits on the left as the bits on the right drop off. It would seem that if you >>33 on an int, then you would be left with all zeroes as all the bits would drop off the right side and the left would be filled in with the sign bit which is zero.
Is there a rule or something that I am not taking into account ?
Any clarification would be helpful.
As always thanks for all the great answers.
Rod
 
Junilu Lacar
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Modulo reduction is applied to the right hand operand based on the width of the left hand operand.
given: l (shift operation) r
If l is an int (32 bits) r is reduced to (r % 32). In your example, (33 % 32) == 1. Or you can think of it as using only the first 5 low order bits of r.
If l is a long (64 bits) r would be reduced to (r % 64), in effect using only the first 6 low order bits of r.
See these threads for more:
http://www.javaranch.com/ubb/Forum24/HTML/010470.html
http://www.javaranch.com/ubb/Forum24/HTML/010607.html

[This message has been edited by JUNILU LACAR (edited July 10, 2001).]
 
Joshua Bloch
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Hi. When you shift an integer with the << or >> operator and the shift distance is greater than or equal to 32, you take the shift distance mod 32 (in other words, you mask off all but the low order 5 bits of the shift distance).
This can be very counterintuitive. For example (i >> 32) == i, for every integer i. You might expect it to shift the entire number off to the right, returning 0 for positive inputs and -1 for negative inputs, but it doesn't; it simply returns i, because (i << (32 & 0x1f)) == (i << 0) == i.
Getting back to your original problem, (i << 33) == (i << (33 & 0x1f)) == (i << 1). You can do the whole thing in binary if you like. You said:
> 270 in binary is :
> 0000 0000 0000 0000 0000 0001 0000 1110
Shifting right by 1, you get:
0000 0000 0000 0000 0000 0000 1000 0111
which is 135.
But a better way to do this problem in your head is to dispense with the binary entirely. The value of i >> s is floor(i / 2<sup>s</sup>) (where s has already been masked off so it's less than 32). So, in your case, 270 << 1 = floor(270/2) = 135. Easy!

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Joshua Bloch
Author of Effective Java
 
Joshua Bloch
Author and "Sun God"
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One note concerning JUNILU LACAR's reply: the % operator is not really a mod operator, it's a remainder operator. The difference is subtle but important. When the first operand is negative, the remainder operator returns a negative result, while a true mod operator always returns a positive result between 0 and (second operand - 1).
In the case of the shift operator's behavior, your really need a true mod operator. For example, (i >> -1) is (i >> 31), because -1 mod 32 is 31. That's why the Java Language Spec describes the shift operator in terms of masking rather than the % operator (Section 15.19).
For what it's worth, java.math.BigInteger provides both a mod method and a remainder method.

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Joshua Bloch
Author of Effective Java
 
Rod Nichols
Greenhorn
Posts: 23
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Junilu and Joshua,
Thank you very much, that clarifies it for me.
Rod
 
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