Left shift but replace the shifted bits with ones-CodePudding

In Python the << operator does the following:

Returns x with the bits shifted to the left by y places (and new bits on the right-hand-side are zeros). This is the same as multiplying x by 2**y.

I want to have another version where it fills 1 on the new bits. Is there a built in function for this, if not how do I write it?

CodePudding user response：

Is x = (x << y) | ((1 << y) - 1) what you're looking for?

First, we shift x left y bits:

x = 21
y = 2
bin(x) == 10101 (in binary)
x << y == 1010100

Then, the number ((1 << y) - 1) gives us a number with only the y lowest bits set to 1, e.g.

1 << 2 == 100
(1 << 2) - 1 == 011

Finally, we or them to get 1010100 | 11 == 1010111.

CodePudding user response：

There's no such function built in, or, for that matter, available in any extension library I know of.

It's pretty easy to do yourself, though. Say the int is n and you want to shift it left by s bits.

First flip all the bits of n (change all 0 bits to 1, and all 1 bits to 0). Then shift left by s. That adds s 0 bits on the right. Then flip the bits again. The new trailing 0 bits are changed to 1 bits, and the original bits of n are restored.

>>> n = 5
>>> bin(n)
'0b101'
>>> ~(~n << 6)
383
>>> bin(_)
'0b101111111'