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

Answer 1

It has to do with signed value math. The >>> pills in zeros for the hight order bits, the >> preservers the sign bit and pulls it in.

Answer 2

some info

the >> operator preserves the leftmost bits. The leftmost bits are filled with the previous content. This is to do with sign extension. In this case there is a 1 at the left and it is preserved. If you do not want to keep the 1 to the left, use the >>> operator which shifts 0's into the leftmost bits

Answer 3

The correct answer has been posted more than once, but not from an authoritative source.

This is from the JLS §15.19 Shift Operators:

The shift operators include left shift <<, signed right shift >>, and unsigned right shift >>>; they are syntactically left-associative (they group left-to-right). The left-hand operand of a shift operator is the value to be shifted; the right-hand operand specifies the shift distance.

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The value of n>>s is n right-shifted s bit positions with sign-extension. The resulting value is ⌊n/2^s⌋. For nonnegative values of n, this is equivalent to truncating integer division, as computed by the integer division operator /, by two to the power s.

The value of n>>>s is n right-shifted s bit positions with zero-extension. If n is positive, then the result is the same as that of n>>s; if n is negative, the result is equal to that of the expression (n>>s)+(2<<~s) if the type of the left-hand operand is int, and to the result of the expression (n>>s)+(2L<<~s) if the type of the left-hand operand is long. The added term (2<<~s) or (2L<<~s) cancels out the propagated sign bit. (Note that, because of the implicit masking of the right-hand operand of a shift operator, ~s as a shift distance is equivalent to 31-s when shifting an int value and to 63-s when shifting a long value.)

Answer 4

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

Answer 5

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 6

From Java Notes: Bitwise Operators:

n >> p (right shift) Shifts the bits of n right p positions. If n is a 2's complement signed number, the sign bit is shifted into the high-order positions.

Example: 5 >> 2 = 1

n >>> p (right shift) Shifts the bits of n right p positions. Zeros are shifted into the high-order positions.

Example: -4 >>> 28 = 15