Python get least significant digits from a float (without using string operations)

Assuming I have the float 12345.6789 and I want to get the six least significant digits (i.e. 45.6789) as an int (i.e. 456789) using bit operations in python (v. 2.6).

How do I do that?

Thanks

PS I do not want to use string operations even if it would be rather easy to: for any float f:

int(str(int(f * 1000))[-10:])

EDIT: This original question is pointless, as shown by comments within. Many 开发者_Go百科apologies... instead methods on getting the least significant digits without using strings are shown below (using ints and modulus)

>>> a = 12345.6789
>>> b = int(a*10000)
>>> b
123456789
>>> c = b % 1000000
>>> c
456789

But - WHY??

Relevant for your string solution is that the float display algorithm changed in python 2.7:

Python 2.6.6 (r266:84292, Sep 15 2010, 15:52:39) 
>>> 12345.6789
12345.678900000001

Python 2.7.0+ (r27:82500, Sep 15 2010, 18:04:55) 
>>> 12345.6789
12345.6789

Whether you use one or the other, there's a problem knowing what, exactly, is the precision of a floating point number. Say your algorithm is given the float 1.23. What's the precision? Maybe it was stored as 1.230, but it just happened to end with a zero digit. So you must know how many digits after the period you wish to preserve.

Also, bit operations do not work on floats:

>>> 12345.6789 << 4
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for <<: 'float' and 'int'

So you must multiply the float by your known precision and use modulo (%), as suggested by other posters.

>>> int(12345.6789*10000)%1000000
456789

but that is not bit operations