Bit masking in Python

Question

I have a byte (from some other vendor) where the potential bit masks are as follows:

value1 = 0x01 value2 = 0x02 value3 = 0x03 value4 = 0x04 value5 = 0x05 value6 = 0x06 value7 = 0x40 value8 = 0x80

I can count on ONE of value1 through value6 being present. And then value7 may or may not be set. value8 may or may not be set.

So this is legal: value2 | value7 | value8 This is not legal: value1 | value3 | value7

I need to figure out whether value 7 is set, value8 is set, and what the remaining value is.

I have the following python code. Is there a more elegant way to do this?

value1 = 0x01
value2 = 0x02
value3 = 0x03
value4 = 0x04
value5 = 0x05
value6 = 0x06
value7 = 0x40
value8 = 0x80

def format_byte_as_bits(value):
    return format(value,'b').zfill(8)

def mask_bits_on_byte(byte,mask):
    inverse_of_mask = mask ^ 0b11111111
    return byte & inverse_of_mask

def parse_byte(byte):

    value7_set = byte & value7 == value7
    value8_set = byte & value8 == value8
    byte = mask_bits_on_byte(byte,value7)
    byte = mask_bits_on_byte(byte,value8)
    base_value = byte
    return value7_set,value8_set,base_value

# Example 1
byte = value3 | value7
value7_set,value8_set,base_value = parse_byte(byte)
print("base_value = "+str(base_value))
print("value7_set = "+str(value7_set))
print("value8_set = "+str(value8_set))
print()

# Output:
# base_value = 3
# value7_set = True
# value8_set = False

# Example 2
byte = value5
value7_set,value8_set,base_value = parse_byte(byte)
print("base_value = "+str(base_value))
print("value7_set = "+str(value7_set))
print("value8_set = "+str(value8_set))
print()

# Output:
# base_value = 5
# value7_set = False
# value8_set = False

# Example 3
byte = value1 | value7 | value8
value7_set,value8_set,base_value = parse_byte(byte)
print("base_value = "+str(base_value))
print("value7_set = "+str(value7_set))
print("value8_set = "+str(value8_set))

# Output:
# base_value = 1
# value7_set = True
# value8_set = True

EDIT - I LOVE stackoverflow. So many useful answers, so quickly! You guys are awesome! Wish I could mark all the answers. But I'll at least give everyone an up vote!

EDIT2 - Based on the answers below, the code is simplified to the following:

value1 = 0x01
value2 = 0x02
value3 = 0x03
value4 = 0x04
value5 = 0x05
value6 = 0x06
value7 = 0x40
value8 = 0x80

def parse_byte(byte):
    return byte & value7, byte & 0x80, byte & 7

# Example 1
byte = value3 | value7
value7_set,value8_set,base_value = parse_byte(byte)
print("base_value = "+str(base_value))
if value7_set: print("value7_set")
if value8_set: print("value8_set")
print()

# Example 2
byte = value5
value7_set,value8_set,base_value = parse_byte(byte)
print("base_value = "+str(base_value))
if value7_set: print("value7_set")
if value8_set: print("value8_set")
print()

# Example 3
byte = value1 | value7 | value8
value7_set,value8_set,base_value = parse_byte(byte)
print("base_value = "+str(base_value))
if value7_set: print("value7_set")
if value8_set: print("value8_set")
print()

Sven Marnach · Accepted Answer · 2011-12-14 17:29:02Z

21

Most of your value* constants aren't actually bit masks, only value7 and value8 are. I'd define another bit mask to extract the lower bits, so I would have three bit masks in total:

mask0 = 0x07
mask1 = 0x40
mask2 = 0x80

Now your function becomes

def parse_byte(byte):
    return byte & mask2, byte & mask1, byte & mask0

I did not convert the results to bool -- I don't see why this should be necessary. When checking the returned value with if, it will be implicitly converted to bool anyway.

Also note that

format(value,'b').zfill(8)

can be simplified to

format(value,'08b')

answered Dec 14, 2011 at 17:29

Sven Marnach

608k123 gold badges966 silver badges865 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Matthew Lund Over a year ago

Very succinct and I appreciate the tip on the zfill as well. I use format(value,'b').zfill(8) all over the place, now I can just use format(value,'08b'). Thanks for your response!

Kat Over a year ago

As an aside, an easy way to make proper bitmasks with nice readable code is to write them like value1 = 1 << 0, value2 = 1 << 1 (etc). That is, take a single bit and just change the shift. Errors are more obvious than with hex or decimal literals. If the mask needs multiple bits set, just | those together (eg, value3 = (1 << 2) | (1 << 3) instead of value3 = 0x0c).

Raymond Hettinger · Accepted Answer · 2011-12-14 17:33:08Z

12

Given a value such as:

>>> x = 0b10001000

You can find out whether the top bits are set with:

>>> bit8 = bool(x & 0b10000000)
>>> bit7 = bool(x & 0b01000000)

To find which lower bit is set, use a dictionary:

>>> bdict = dict((1<<i, i+1) for i in range(6))
>>> bdict[x & 0b00111111]
4

answered Dec 14, 2011 at 17:33

Raymond Hettinger

229k67 gold badges405 silver badges504 bronze badges

Comments

stew · Accepted Answer · 2011-12-14 17:24:54Z

4

You don't need the other two functions:

def parse_byte(byte):
    value7_set = byte & value7 == value7
    value8_set = byte & value8 == value8
    base_value =  byte & 7
    return value7_set,value8_set,base_value

answered Dec 14, 2011 at 17:24

stew

11.4k39 silver badges49 bronze badges

1 Comment

Matthew Lund Over a year ago

Thanks! Didn't think of the byte & 7.

Mark Ransom · Accepted Answer · 2011-12-14 17:27:43Z

3

It's a little verbose but perfectly fine. The only change I'd make is to simplify parse_byte:

def parse_byte(byte):

     value7_set = byte & value7 == value7
     value8_set = byte & value8 == value8
     base_value = mask_bits_on_byte(byte,value7 | value8)
     return value7_set,value8_set,base_value

answered Dec 14, 2011 at 17:27

Mark Ransom

310k44 gold badges423 silver badges660 bronze badges

Comments

MEMYANDi · Accepted Answer · 2023-03-31 08:40:22Z

-1

Wlio to run the byte in the register and use the instruction to get a result will also verify data and cc

answered Mar 31, 2023 at 8:40

MEMYANDi

1

1 Comment

Community Over a year ago

As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.

Collectives™ on Stack Overflow

Bit masking in Python

5 Answers 5

2 Comments

Comments

1 Comment

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

2 Comments

Comments

1 Comment

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related