Python Bit Shifting [closed]

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Assume I have a 1 source variable (int) which is 10 bits long. Let's call it src. I have 2 destination variables (int) which are 8 bits long namely byte1 and byte2.

       MSB               LSB 
src    0 1 2 3 4 5 6 7 8 9  --- bit numbers
       1 1 0 0 1 0 1 1 1 0  --- assigned bits
       value 814

What I would like to do is copy the first 8 bits from src to byte1 and the last 2 bits from src to the MSB of byte2 so byte1 and byte2 should contain the following:

       MSB           LSB 
byte1  0 1 2 3 4 5 6 7   --- bit numbers
       1 1 0 0 1 0 1 1
       value 203

I can do this with byte1 = src >> 2 which works perfectly as intended

       MSB           LSB 
byte2  0 1 2 3 4 5 6 7
       1 0 0 0 0 0 0 0
       value 128

Where I am stuck is with byte2. I have tried creating a mask for the last 2 bits and ANDing the source and mask together and then ORing the extracted bits.

Include the actual code you currently have for byte2 rather than words. And what is your actual result for byte2 . — Computable
– Computable, Commented Sep 17 at 11:28

Kelly Bundy · Accepted Answer · 2025-09-17 12:23:37Z

3

byte1, byte2 = (src << 6).to_bytes(2)

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answered Sep 17 at 12:23

Kelly Bundy

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Nixxen · Accepted Answer · 2025-09-17 11:29:27Z

2

You can do (src & 0b11) to get the last two bits of the src, then shift those two to using << 6, placing the bits at the MSB side.
Python has arbitrary precision, so, as far as I know, truncation doesn't matter. In other languages, it might also be good practice to explicitly truncate your result using & 0xFF.

answered Sep 17 at 11:29

Nixxen

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jackal · Accepted Answer · 2025-09-17 12:22:04Z

2

Shifting right by 2 is correct for getting your byte1 value. You could also floor divide by 4.

For byte2, you can "mask" src using a bitwise AND to get the 2 least significant bits. Then shift left by 6 or multiply by 64 like this:

src = 814

assert src < 1024 # ensure no more than 10 bits in src

w = 8

print(f"{src=:0{w}b}")

byte1a = src >> 2 # right shift by 2
print(f"{byte1a=:0{w}b}")

byte1b = src // 4 # floor divide by 4
print(f"{byte1b=:0{w}b}")

byte2a = (src & 3) << 6 # left shift by 6
print(f"{byte2a=:0{w}b}")

byte2b = (src & 3) * 64 # multiply by 64
print(f"{byte2b=:0{w}b}")

Output:

src=1100101110
byte1a=11001011
byte1b=11001011
byte2a=10000000
byte2b=10000000

edited Sep 17 at 12:22

answered Sep 17 at 12:11

jackal

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