CODEWITHSUNDEEP
javabit-shift
Suresh
Suresh

Reputation: 31

java asymmetry in right and left shift operators

Consider the following code.

  1. 1 >> 34 = 0,
  2. 1 << 34 = 4 // circular shift

why such asymmetry ?

Upvotes: 0

Views: 361

Answers (3)

Rohit Jain
Rohit Jain

Reputation: 213213

From JLS §15.19:

If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x1f (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive.

So, it's not that << is circular and >> is not. It's just the value you chose, that made you believe that.

With shifting 1 by 34, considering the lowest 5 bits:

1 >> 34  === 1 >> 2 === 0
1 << 34  === 1 << 2 === 4

Since 34 & 0x1F == 2.

Now, take some larger value, for which bit shifting by 2 will not get you 0 technically:

4 >> 34 == 4 >> 2 == 1
4 << 34 == 4 << 34 == 16

The shift distance for int type are always calculated taking the 5 lowest-order bits. It's not a circular shift operation. Neither of them are. This is how they are intended to behave.

Upvotes: 5

Peter Lawrey
Peter Lawrey

Reputation: 533472

Shift only uses the lower 5 bits for an int and the lower 6 bit for a long.

1 >> 34 = 0

This is the same as 1 >> 2 which is 0 as this is not circular rotation.

1 << 34 = 4 // circular shift

This is the same as 1 << 2 as the shift only uses the lower 5 bits for an int and the lower 6 bit for a long.

This is not circular shifting.

e.g.

Integer.MIN_VALUE << 1 // is 0

Upvotes: 2

Theolodis
Theolodis

Reputation: 5102

When shifting right with the signed shift operator 1>>34, following the oracle documentation, the left is filled up with the sign, in this case, as 1 is positive, the sign is 0. The one gets lost with the first shift 1>>1 = 0

When shifting left with the signed shift operator 1<<34 the following happens:

0x0001<<34 = (0x0001 << 32) << 2 = 0x0004 = 4

Upvotes: 0

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