Shifting by Negative Numbers

Question

I am pretty confused with this expression here. I am a Java programmer but I am not very well versed with bit manipulation.

I think I understand the below correctly:

Input : 1 << 10
Output: 0000000000000000000010000000000

For positive numbers, I think it is you move 1 by 10 bits.

The confusion is when I have the below:

int val = -10 (binary representation : 1111111111111111111111111110110 )
Input : 1 << val
Output: 0000000010000000000000000000000 

That would be really great if someone can explain me the meaning of left shifting or right shifting by negative number.

Answer 1

<< (and other shift operators) only takes 5 least significant bits of its right operand for int, and 6 for long, because it makes no sense to shift int by more than 31.

In your case it's 0b10110 = 22.

Therefore 1 << (-10) is equivalent to 1 << 22.

Answer 2

From the JLS, 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.

In other words,

1 << -10

is equivalent to:

1 << (-10 & 0x1f)

... which is

1 << 22