bitwise >>> strange behaviour

[Nancy Antony]

Hi Ranchers,

when I write int i=-1;

i=i>>>31;
System.out.println(i); // i is 1 understood

when I write

int i=-1;
i=i>>>32;
System.out.println(i); // value is -1

Why so?

Regards,
Nancy

[Stephan van Hulst]

All bitwise shift operators use modulo 32 shifting, except if the left operand is a long, then it uses modulo 64 shifting.

Simply said, if you shift your operand by 32 bits, you shift it by 32 % 32 == 0, so you don't shift it at all.

Also, on a slightly related note, you shouldn't use the >>> operator on shorts or bytes, because it can lead to unexpected results.

[Ankit Garg]

Nancy remember that bitwise operators are not on the exam anymore...

[Javin Paul]

Hi Stephan,

Stephan van Hulst wrote:
Also, on a slightly related note, you shouldn't use the >>> operator on shorts or bytes, because it can lead to unexpected results.

Can you please elaborate this point , I know that >>> (right shift without sign) doesn't filled left most bit with sign bit instead it just fill that with zero.

and as per my knowledge byte ans short are unsigned so eventually it fill zero only but how could it lead to unexpected result ?

Thanks
Javin

[Stephan van Hulst]

Think about it. What happens to operands of types byte and short before operations are carried out? They get promoted to int.

Here is an illustration. For the sake of brevity, let's say that ints are 16 bits, and shorts are 8 bits.

short i = -8; // 11111000
i >>> 5;      // 00000111 ?

Nope. The short is sign extended to int before the operation:

(int) i; // 11111111 11111000
i >>> 5; // 00000111 11111111

This problem gets worse if you use an unsigned right shift compound assignment operator.

i >>>= 5; // 11111111

[Javin Paul]

Thanks Stephan ,its pretty clear now.