Right Shift in java

Question

I am just stuck with this small logic that i am not getting it right

int is 32 bits so suppose taking 20 in binary would be like

// 00000000000000000000000000010100

.. now if I perform rightshift operation say 4

int a = 20>>4;   
// 00000000000000000000000000000001

..so result is 1

Now say I again take 20 and do 5 right shift operation

int b = 20>>5;  //00000000000000000000000000000000

..so result is 0

Now if I do 32 right shift ... why I am getting the same number as I assigned.??

int c = 20>>32; //how does this prints 20 again??

System.out.println("right shift 4= "+a+"\n right shift 5= "+b+"right shift 32 = "+c);

So what I was expecting is after 5 shifts ..any number of shifts should have resulted in the result 0.. but on 32 shifts why I am getting back the assigned value?

A sane compiler would have warned you.

devnull
– devnull

2014-03-08 09:06:40 +00:00
Commented Mar 8, 2014 at 9:06 — devnull
– devnull, Commented Mar 8, 2014 at 9:06

Erwin Bolwidt · Accepted Answer · 2014-03-08 09:26:38Z

7

When you shift an int only the lower 5 bits of the shift are used. When you shift a long only the lowest 6 bits are used.

From the Java Language Specification section 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.

If the promoted type of the left-hand operand is long, then only the six 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 0x3f (0b111111). The shift distance actually used is therefore always in the range 0 to 63, inclusive.

edited Mar 8, 2014 at 9:26

Erwin Bolwidt

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answered Mar 8, 2014 at 9:05

Peter Lawrey

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1 Comment

Pshemo Over a year ago

While this is true it would be nice to explain it bit more. What this answer is trying to say is that since 32 in bit representation is 100000 last 5 bits are 00000 which represents 0 so there will be no bit shifting. For long type no shifting would happen for 64 (1000000) and its multiplications.

Evgeniy Dorofeev · Accepted Answer · 2014-03-08 09:17:24Z

JLS 15.19. Shift Operators

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.

If the promoted type of the left-hand operand is long, then only the six 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 0x3f (0b111111). The shift distance actually used is therefore always in the range 0 to 63, inclusive

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