How to modify multiple bits in an integer?

Question

Challenge:

I would like to write a method which would take

INTEGER value
START_BIT
Number of bits to be replaced after the START_BIT
REPLACING_BIT (boolean)

Return: the modified value based on the inputs.

Input Example:

Decimal: 255 or in Binary: 11111111

Expected output:

Decimal: 227 or in Binary: 11100011

i.e 3 bits are replaced with ZEROS starting from second bit position.

Current code:

def set_bit(original_int:int, start_bit_pos:int, number_of_bits:int, bit_value:bool):
    if number_of_bits < 1: number_of_bits = 1
        
    out = original_int
    for i in range(number_of_bits):
        if bit_value == 1:
            out = out | (1<<(start_bit_pos+i))
        else:
            out = out & ~(1<<(start_bit_pos+i))
    print('old binary: '+ '{:b}'.format(original_int),f' Decimal: {original_int}' )
    print('new binary: '+ '{:b}'.format(out), f' Decimal: {out}' )
    return out

set_bit(original_int = 255, start_bit_pos= 2, number_of_bits = 3, bit_value = 0)
set_bit(original_int = 64, start_bit_pos= 2, number_of_bits = 3, bit_value = 1)

Current issue:

The above code is functional, however I do feel it is an overkill, so I wonder if there is a standard python function which could do a similar thing?

Red · Accepted Answer · 2021-01-08 01:29:31Z

3

A sequence of 1 bits is 1.
A sequence of 2 bits is 11 or decimal 3.
A sequence of 3 bits is 111 or decimal 7.
A sequence of N bits is 2^N-1, which is the same as (1 << N) - 1.

You have to also shift that to the left if the starting position is not 0.

Therefore, you can simply do this:

def set_bit(original_int:int, start_bit_pos:int,
            number_of_bits:int, bit_value:bool):
    mask = (1 << number_of_bits) - 1 << start_bit_pos
    if bit_value:
        return original_int | mask
    return original_int & ~mask

edited Jan 8, 2021 at 1:29

Red

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answered Jan 8, 2021 at 1:03

zvone

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Comments

filipmu · Accepted Answer · 2021-03-25 02:25:57Z

Here is a function that can update an arbitrary number of bits at an arbitrary location in an arbitrary word size. Here x is the original integer, start_bit is the location of first insertion, starting from 0 from the right (least significant bit), field_size is the number of bits to replace, and val is the value to plug into the field. word_size should reflect the number of bits expected in x

def apply_bits(x, start_bit, field_size, val, word_size=8):
    
    print("x=",bin(x))
    
    mask = 2**word_size - 1 - (2**field_size -1) * (2**start_bit)
    print("mask=",bin(mask))
    
    val = min(val, 2**(field_size) - 1)
    print("val=", bin(val))
    
    w = val * 2**start_bit
    
    print("w=",bin(w))
        
    y = (x & mask) | w
    print("y=",bin(y))
    
    return(y)

for example:

>>> apply_bits(x=0b10000001,start_bit=2,field_size=4,val=0b1010,word_size=8)
   x= 0b10000001
 val= 0b1010
mask= 0b11000011
   w= 0b101000
   y= 0b10101001

and

apply_bits(x=0b10101010101010101,start_bit=4,field_size=8,val=0b11100111,word_size=16)
   x= 0b10101010101010101
 val= 0b11100111
mask= 0b1111000000001111
   w= 0b111001110000
   y= 0b101111001110101

First the function makes a mask of zeros at the location of the field. Then the value is constrained to fit into the field size and shifted into position. Finally the field is zero-ed out using bitwise-and, and the new values are put in using bitwise-or. The print statements let you follow along the process.

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