How I do circular shifting (rotation) of int in Python?

Question

In Java right rotation is done using:

 (bits >>> k) | (bits << (Integer.SIZE - k))

But how to do similar thing in Python?

I tried to do (as described here):

n = 13
d = 2
INT_BITS = 4
print(bin(n))
print(bin((n >> d)|(n << (INT_BITS - d)) & 0xFFFFFFFF))

Output:

0b1101
0b110111

But I could not interpret this as a right rotation.

Also is it possible to perform the rotation by excluding leading zeroes, for example:

rightRotation of (...0001101) = 1110 not 1000...110

Revisit the site, you miss the `& 0xFFFFFFFF` part for the right rotation. — AnsFourtyTwo, Dec 03 '19 at 09:08
Thank you, @SimonFink I edited my question, but still produce wrong output. — adnanmuttaleb, Dec 03 '19 at 09:16
It is a mistake, it should be '0xF' for a four number of bit, now it works. — adnanmuttaleb, Dec 03 '19 at 09:18

score 0 · Answer 1 · answered Dec 03 '19 at 09:23

0

It is my mistake, if you want to change INT_BITS to 4 you also need to change 0xFFFFFFFF to 0xF (one hex equals 4-bits):

n = 13
d = 2
INT_BITS = 4
print(bin(n))
print(bin((n >> d)|(n << (INT_BITS - d)) & 0xF))

will output:

0b1101
0b111

answered Dec 03 '19 at 09:23

adnanmuttaleb

1 Answers1