Java's >> versus >>> Operator? [duplicate]

Question

I'm without my Java reference book and I'm having a tough time finding an answer with Google.

What is the difference between the ">>" and ">>>" operators in Java?

int value = 0x0100;

int result = (value >> 8);
System.out.println("(value >> 8) = " + result);  // Prints: "(value >> 8) = 1"

result = (value >>> 8);
System.out.println("(value >>> 8) = " + result); // Prints: "(value >>> 8) = 1"

Answer 1

Signed integers use the high-order bit to denote sign.

So >> preserves the sign, while >>> doesn't. This is why >> is referred to as the arithmetic shift and >>> is the logical shift.

This way, you can do (assuming 32-bit integers) the following:

• -10 >> 1 yields -5 (0xFFFFFFF6 >> 1 yields 0xFFFFFFFB - notice the high-order bit stays the same.)
• -10 >>> 1 yields 2147483643 (0xFFFFFFF6 >>> 1 yields 0x7FFFFFFB - notice all of the bits were shifted, so the high-order bit is now zero. The number is no longer negative according to twos-complement arithemetic.)

For positive integers, >> and >>> act the same, since the high-order bit is already zero.

It also explains why there is no need for a <<< operator. Since the sign would be trashed by sliding the bits to the left, it would map to no reasonable arithmetic operation.

Answer 2

>>> is logical shift, >> is arithmetic shift.