Log base 2 of large integers

[Steven D'Aprano]

Mok-Kong Shen wrote:

> 
> I like to compute log base 2 of a fairly large integer n but
> with math.log(n,2) I got:
> 
> OverflowError: long int too large to convert to float.
> 
> Is there any feasible work-around for that?

If you want the integer log2, that is, the floor of log2, the simplest way
is calculate it like this:

def log2(n):
    """Return the floor of log2(n)."""
    if n <= 0: raise ValueError
    i = -1
    while n:
        n //= 2
        i += 1
    return i

log2(511)
=> returns 8
log2(512)
=> returns 9
log2(513)
=> returns 9


Does that help?



-- 
Steven