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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T05:32:18+00:00 2026-05-14T05:32:18+00:00

There is a byte at a specific index in a byte string which represents

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There is a byte at a specific index in a byte string which represents eight flags; one flag per bit in the byte. If a flag is set, its corresponding bit is 1, otherwise its 0. For example, if I’ve got

b'\x21'

the flags would be

0001 0101    # Three flags are set at indexes 0, 2 and 4
             # and the others are not set

What would be the best way to get each bit value in that byte, so I know whether a particular flag is set or not? (Preferably using bitwise operations)

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  1. Editorial Team
    Editorial Team
    2026-05-14T05:32:18+00:00Added an answer on May 14, 2026 at 5:32 am

    Typically, the least-significant bit is bit index 0 and the most-significant bit is bit index 7. Using this terminology, we can determine if bit index k is set by taking the bitwise-and with 1 shifted to the left by k. If the bitwise and is non-zero, then that means that index k has a 1; otherwise, index k has a 0. So:

    def get_bit(byteval,idx):
        return ((byteval&(1<<idx))!=0);
    

    This will correctly determine the value of bits at indices 0…7 of the byte, going from right-to-left (i.e. the least significant bit to the most significant bit, or equivalently from the 1s place to the 27 = 128 place).

    Why it works
    I figured I should add an explanation of why it works…

    1<<0 is 1 = 0000 0001
    1<<1 is 2 = 0000 0010
    1<<2 is 4 = 0000 0100

    As you can see, 1<<k is equivalent to 2k and contains a 1 at exactly the index we are interested and at no other location. Consequently, the bitwise and with 1<<k will either return 0 or 1<<k; it will be 0 if the bit at the index we are interested in is 0 (because 1 and 0 is 0, and all other bits in 1<<k are zero). If the bit we are interested in is 1, then we get a 1 and a 1 in that position, and a 0 and something else, everywhere else.

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