confusion in left shift operator in python

Question

I am trying to use left shift operator on the 16 bit binary representation of a integer

Code written is below:

def showbits(x):
    return bin(x)[2:].zfill(16)
i=5225
print "Decimal %d is same as binary" % (i)
print showbits(i)
for j in range(0,5,1):
    k=i<<j
    print "%d right shift % gives" % (i,j)
    print showbits(k)

Output:

Decimal 5225 is same as binary
0001010001101001
5225 right shift  0ives
0001010001101001
5225 right shift  1ives
0010100011010010
5225 right shift  2ives
0101000110100100
5225 right shift  3ives
1010001101001000
5225 right shift  4ives
10100011010010000

The main problem is that when it is shifting the leading '1', it is not vanishing but it is increasing one more bit...

Any solution for that?

Answer 1

You'd mask the resulting value, with & bitwise AND:

mask = 2 ** 16 - 1
k = (i << j) & mask

Here 16 is your desired bit width; you could use i.bit_length() to limit it to the minimum required size of i, but that'd mean that any left shift would drop bits.

The mask forms a series of 1 bits the same width as the original value; the & operation sets any bits to 0 outside of these:

>>> 0b1010 & 0b111
2
>>> format(0b1010 & 0b111, '04b')
'0010'

Some side notes:

• You are left shifting, not right shifting.

• You appear to have forgotten to a d in your debug print:

  print "%d left shift %d gives" % (i,j)
  

  There was a lone % there that combined with the g for gives to make %g (floating point formatting).

• You can use:

  def showbits(x):
      return format(x, '016b')
  

  to format an integer to a 0-padded 16-character wide binary representation without the 0b prefix.