Unsigned right shift operator in Java

By: Baski Viewed: 170 times  Printer Friendly Format    


The >> operator automatically fills the high-order bit with its previous contents each time a shift occurs. This preserves the sign of the value. However, sometimes this is undesirable. For example, if you are shifting something that does not represent a numeric value, you may not want sign extension to take place. This situation is common when you are working with pixel-based values and graphics. In these cases you will generally want to shift a zero into the high-order bit no matter what its
initial value was. This is known as an unsigned shift. To accomplish this, you will use Java's unsigned, shift-right operator, >>>, which always shifts zeros into the high-order bit.

The following code fragment demonstrates the >>>. Here, a is set to -1, which sets all 32 bits to 1 in binary. This value is then shifted right 24 bits, filling the top 24 bits with zeros, ignoring normal sign extension. This sets a to 255.

int a = -1;
a = a >>> 24;

Here is the same operation in binary form to further illustrate what is happening:

11111111 11111111 11111111 11111111     -1    in binary as an int
>>>24
00000000 00000000 00000000 11111111     255 in binary as an int

The >>> operator is often not as useful as you might like, since it is only meaningful for 32- and 64-bit values. Remember, smaller values are automatically promoted to int in expressions. This means that sign-extension occurs and that the shift will take place on a 32-bit rather than on an 8- or 16-bit value. That is, one might expect an unsigned right shift on a byte value to zero-fill beginning at bit 7. But this is not the case, since it is a 32- bit value that is actually being shifted. The following program demonstrates this effect:

// Unsigned shifting a byte value.
class ByteUShift {
static public void main(String args[]) {
char hex[] = {
'0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', 'a', 'b', 'c', 'd', 'e', 'f'
};
byte b = (byte) 0xf1;
byte c = (byte) (b >> 4);
byte d = (byte) (b >>> 4);
byte e = (byte) ((b & 0xff) >> 4);
System.out.println(" b = 0x"
+ hex[(b >> 4) & 0x0f] + hex[b & 0x0f]);
System.out.println(" b >> 4 = 0x"
+ hex[(c >> 4) & 0x0f] + hex[c & 0x0f]);
System.out.println(" b >>> 4 = 0x"
+ hex[(d >> 4) & 0x0f] + hex[d & 0x0f]);
System.out.println("(b & 0xff) >> 4 = 0x"
+ hex[(e >> 4) & 0x0f] + hex[e & 0x0f]);
}
}

The following output of this program shows how the >>> operator appears to do nothing when dealing with bytes. The variable b is set to an arbitrary negative byte value for this demonstration. Then c is assigned the byte value of b shifted right by four, which is 0xff because of the expected sign extension. Then d is assigned the byte value of b unsigned shifted right by four, which you might have expected to be 0x0f, but is actually 0xff because of the sign extension that happened when b was promoted to int before the shift. The last expression sets e to the byte value of b masked to 8 bits using the AND operator, then shifted right by four, which produces the expected value of 0x0f. Notice that the unsigned shift right operator was not used for d, since the state of the sign bit
after the AND was known.

b = 0xf1
b >> 4 = 0xff
b >>> 4 = 0xff
(b & 0xff) >> 4 = 0x0f



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