Python: Extracting bits from a byte [duplicate]

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Using Python How can I read the bits in a byte? (11 answers)

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I'm reading a binary file in python and the documentation for the file format says:

Flag (in binary)Meaning

1 nnn nnnn Indicates that there is one data byte to follow that is to be duplicated nnn nnnn (127 maximum) times.

0 nnn nnnn Indicates that there are nnn nnnn bytes of image data to follow (127 bytes maximum) and that there are no duplications.

n 000 0000 End of line field. Indicates the end of a line record. The value of n may be either zero or one. Note that the end of line field is required and that it is reflected in the length of line record field mentioned above.

When reading the file I'm expecting the byte I'm at to return 1 nnn nnnn where the nnn nnnn part should be 50.

I've been able to do this using the following:

flag = byte >> 7
numbytes = int(bin(byte)[3:], 2)

But the numbytes calculation feels like a cheap workaround.

Can I do more bit math to accomplish the calculation of numbytes?

How would you approach this?

• jfs: related: Using Python How can I read the bits in a byte?

Answer 1

The classic approach of checking whether a bit is set, is to use binary "and" operator, i.e.

x = 10 # 1010 in binary
if x & 0b10: # explicitly: x & 0b0010 != 0
 print('First bit is set')

To check, whether n^th bit is set, use the power of two, or better bit shifting

def is_set(x, n):
 return x & 2 ** n != 0 
 # a more bitwise- and performance-friendly version:
 return x & 1 << n != 0
is_set(10, 1) # 1 i.e. first bit - as the count starts at 0-th bit
>>> True

• Evan Borgstrom: Thanks, but that doesn't really answer my question. Given byte = 178, how would you extract flag == 1 & numbytes == 50?
• tom10: +1: but one doesn't need the !=0 part, if x&0b10 should be sufficient.
• Sven Marnach: != 0 is not quite as bad as == True, but it is close. :)
• D.Shawley: Explicit is better than implicit so x & 0b10 != 0 is fine by me
• Wolf: print('First bit is set') – the word first is very confusing here.

Answer 2

You can strip off the leading bit using a mask ANDed with a byte from file. That will leave you with the value of the remaining bits:

mask = 0b01111111
byte_from_file = 0b10101010
value = mask & byte_from_file
print bin(value)
>> 0b101010
print value
>> 42

I find the binary numbers easier to understand than hex when doing bit-masking.

EDIT: Slightly more complete example for your use case:

LEADING_BIT_MASK = 0b10000000
VALUE_MASK = 0b01111111
values = [0b10101010, 0b01010101, 0b0000000, 0b10000000]
for v in values:
 value = v & VALUE_MASK
 has_leading_bit = v & LEADING_BIT_MASK
 if value == 0:
 print "EOL"
 elif has_leading_bit:
 print "leading one", value
 elif not has_leading_bit:
 print "leading zero", value

• Evan Borgstrom: Thanks. I too prefer the binary numbers in this case.
• jfs: @EvanBorgstrom On Python 3, you could write: data = b'\xaa\x55\x00\x80' and for byte in data: .... Don't use bytes name, it is a builtin type.

Answer 3

Instead of int(bin(byte)[3:], 2), you could simply use: int(bin(byte>>1),2)

Answer 4

If I read your description correctly:

if (byte & 0x80) != 0:
 num_bytes = byte & 0x7F

Answer 5

there you go:

class ControlWord(object):
 """Helper class to deal with control words.
 Bit setting and checking methods are implemented.
 """
 def __init__(self, value = 0):
 self.value = int(value)
 def set_bit(self, bit):
 self.value |= bit
 def check_bit(self, bit):
 return self.value & bit != 0
 def clear_bit(self, bit): 
 self.value &= ~bit

Answer 6

not sure I got you correctly, but if I did, this should do the trick:

>>> x = 154 #just an example
>>> flag = x >> 1
>>> flag
1
>>> nb = x & 127
>>> nb
26

Answer 7

You can do it like this:

def GetVal(b):
 # mask off the most significant bit, see if it's set
 flag = b & 0x80 == 0x80
 # then look at the lower 7 bits in the byte.
 count = b & 0x7f
 # return a tuple indicating the state of the high bit, and the 
 # remaining integer value without the high bit.
 return (flag, count)
>>> testVal = 50 + 0x80
>>> GetVal(testVal)
(True, 50)

Answer 8

You can use ReverseBox
For example if you want to get 3 bits on position 5 from number 2273, you can write something like this:

from reversebox.io_files.bytes_helper_functions import get_bits
result = get_bits(2273, 3, 5)
print(result)

Result:

7

• tripleee: Please don't post the same answer to multiple questions. Are you affiliated with the site you are promoting?