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int x = -50;
x = x + 2;
x >>>>>= 2;

The >>>= operator is a logical right shift; it fills the leftmost bits with zeroes, regardless of the number's original sign. Using logical right shift for division produces an incorrect result when the dividend (x in this example) contains a negative value. Arithmetic right shift (the >>= operator) can replace division, as shown in the following example:

int x = -50;
x = x + 2;
x >>>>>= 2;

However, the result is less maintainable than making the division explicit. Refer to guideline NUM04-J. Use shift operators correctly for examples of sign extension with respect to the shift operators.

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// b[] is a byte array, initialized to 0xff
byte[4] b = new byte[] {-1, -1, -1, -1};
int result = 0;
for (int i = 0; i < 4; i++) {
  result = ((result << 8) + b[i]);
}

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byte[4] b = new byte[] {-1, -1, -1, -1};
int result = 0;
for (int i = 0; i < 4; i++) {
  result = ((result << 8) + (b[i] & 0xff));
}

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byte[4] b = new byte[] {-1, -1, -1, -1};
int result = 0;
for (int i = 0; i < 4; i++) {
  result = ((result << 8) | (b[i] & 0xff));
}

Exceptions

INT03-EX1EX0: Bitwise operations may be used to construct constant expressions.

int limit = 1 << 17 - 1; // 2^17 - 1 = 131071

INT03-EX2EX1: Data that is normally treated arithmetically may be treated with bitwise operations for the purpose of serialization or deserialization. This is often required for reading or writing the data from a file or network socket, It may also be used when reading or writing the data from a tightly packed data structure of bytes.

int value = /* interesting value */
Byte[] bytes[] = new Byte[4];
for (int i = 0; i < bytes.length; i++) {
  bytes[i] = value >> (i*8) & 0xFF;
}
/* bytes[] now has same bit representation as value  */

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