问题
I came across the question "Why is -1 zero fill right shift 1=2147483647 for integers in Java?"
I understood the concept of zero fill right shift perfectly well from the above question's answer. But when I tried to find -1>>1, I am getting a totally complex answer which I felt difficult to understand.
-1 in binary form is as follows: 11111111111111111111111111111111 After flipping the bits, I got: 00000000000000000000000000000000 Upon adding 1 to it, I got: 00000000000000000000000000000001 Now shifting one position right: 00000000000000000000000000000000 After flipping the bits, I got: 11111111111111111111111111111111 Now adding 1 to it: 00000000000000000000000000000000
I don't understand how -1>>1 is -1 itself, then?
回答1:
When you do a normal right-shift (i.e. using >>, also known as an arithmetic right shift, as opposed to >>>, which is a logical right shift), the number is sign extended.
How this works is as follows:
When we right-shift we get an empty spot in front of the number, like so:
11111111111111111111111111111111
?1111111111111111111111111111111(1) (right-shift it one place)
The last 1 is shifted out, and in comes the ?.
Now, how we fill in the ? is dependent on how we shift.
If we do a logical shift (i.e. >>>), we simply fill it with 0.
If we do a arithmetic shift (i.e. >>), we fill it with the first bit from the original number, i.e. the sign bit (since it's 1 if the number is negative, and 0 if not). This is called sign extension.
So, in this case, -1 >> 1 sign-extends the 1 into the ?, leaving the original -1.
Further reading:
- Arithmetic shift on Wikipedia
- Logical shift on Wikipedia
回答2:
>> is a 'signed right shift' operator, so it will preserve the leftmost bit.
So as -1 (dec) is 0xb11111111111111111111111111111111 (bin), right shifting it using >> will preserve the 1 and the result is again 0xb11111111111111111111111111111111 (bin).
For unsigned right shift, the operator >>> can be used. It always fills up with zeroes (0). In your case, -1 >>> 1 results in 0b01111111111111111111111111111111, or 2147483647, which is the largest possible (positive) 32 bit two's complement (and equal to Integer.MAX_VALUE).
来源：https://stackoverflow.com/questions/20468564/why-is-1-right-shift-1-1-in-java