How to modify bits in an integer?

These work for integers of any size, even greater than 32 bit:

def set_bit(value, bit):    return value | (1<<bit)def clear_bit(value, bit):    return value & ~(1<<bit)

If you like things short, you can just use:

>>> val = 0b111>>> val |= (1<<3)>>> '{:b}'.format(val)'1111'>>> val &=~ (1<<1)'1101'

You just need:

def set_bit(v, index, x):  """Set the index:th bit of v to 1 if x is truthy, else to 0, and return the new value."""  mask = 1 << index   # Compute mask, an integer with just bit 'index' set.  v &= ~mask          # Clear the bit indicated by the mask (if x is False)  if x:    v |= mask         # If x was True, set the bit indicated by the mask.  return v            # Return the result, we're done.>>> set_bit(7, 3, 1)15>>> set_bit(set_bit(7, 1, 0), 3, 1)13

Note that bit numbers (index) are from 0, with 0 being the least significant bit.

Also note that the new value is returned, there's no way to modify an integer "in place" like you show (at least I don't think so).

You can use bitwise opertions.http://wiki.python.org/moin/BitwiseOperators

if you want to set a given bit to 1 you can use bitwise 'or' with 1 on given position:

0b00000111 | 0b00001000 = 0b00001111

to set a given bit to 0 you can use bitwise 'and'

0b00001111 & 0b11111011 = 0b00001011

Note that 0b prefix is for binary numbers and 0x is for hexadecimal.