shift : Java Glossary

2008-03-12 by Roedy Green ©1996-2008 Canadian Mind Products
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shift

Java has a number of operators for multiplying/dividing by powers of two, i.e. shifting left or right by bits. These are very fast operations, much faster than the using multiplication, division and Math.pow.

Shift Operators
Operation	Operator	Example	Effect
left shift	<<	0x0a << 4 ⇒ 0xa0	Shift left n bits, multiply by 2ⁿ
logical right shift	>>>	0xa0 >>> 4 ⇒ 0x0a 0xffffffa0 >>> 4 ⇒ 0x7ffffffa	unsigned shift right n bits, 0 bit fills in on the left, divide by 2ⁿ
arithmetic right shift	>>	0xa0 >> 4 ⇒ 0x0a 0xffffffa0 >> 4 ⇒ 0xfffffffa	signed shift right n bits, sign bit fills in on the left, divide by 2ⁿ
left shift	<<=	x <<= 4	shorthand for x = x << 4
logical right shift	>>>=	x >>>= 4	shorthand for x = x >>> 4
arithmetic right shift	>>=	x >>= 4	shorthand for x = x >> 4

Creating Masks Dynamically

You can create powers of two, (one-bit masks) by shifting 1 left a variable amount, e.g. (1<<n). Normally you create masks with hex literals, e.g. 0x007f. To dynamically create a mask with n low order 1’s, use (1<<n)-1. Beware, for int this will not work for n=32 since shifts are done modulo 32.

Shiftless Shift

The shift operators can give some surprising results. You can’t shift ints by more than 31 bits. If you try it, you will shift by that amount modulo 32. Similarly you can’t shift longs my more than 63 bits. If you try it, you will shift by that amount modulo 64. This applies to <<< << and >> So:

int i = -1;
i = i >>> 32;
// i is -1, not 0 as you might expect.

long k = 1;
k = k << 65;
// k is 2, not 0 as you might expect.

As a corollary, shifting left 32 bits does not clear an int. It is a no-op! Be especially careful when working with ints that will eventually be promoted to longs.

Be especially careful with any calculated shift operands. The shift modulo behavior makes creation of bit selections masks unnecessarily tricky. Suppose you want a general way of extracting bits i through j of an int. An obvious way to do it is to right shift the input by j bits, then & it with a mask containing j-i+1 one bits at the low order end. An obvious way to construct an n-bit right justified mask is (1<<n)-1. This is not correct in Java for n==32. (1 <<32) is 1, not 0, so subtracting 1 from it will result in a zero mask rather than an all-ones mask.

Java works this way for speed since many CPUs work that way.

Sun RFE number 4495754 : Shift fixed for n==32

In the meantime here is a little class you can use:

binary
bit
boolean
division
masking
Math.pow
multiplication
precedence

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