When is divmod(a,b)[0] == floor(a/b)-1 ?

[Robert Kern]

On 2009-09-24 14:40 PM, kj wrote:
>
> The docs for divmod include the following:
>
>      divmod(a, b)
>      ...For floating point numbers the result is (q, a % b), where q
>      is usually math.floor(a / b) but may be 1 less than that. ...
>
> I know that floating point math can sometimes produce "unexpected"
> results, so the above caveat is not entirely surprising.  Still,
> I would find it helpful to see a specific example where
> divmod(a, b)[0] is equal to math.floor(a/b)-1.  Does anybody know
> one?

In [21]: a = 10.0

In [22]: b = 10.0 / 3.0

In [24]: divmod(a, b)[0]
Out[24]: 2.0

In [25]: math.floor(a / b) - 1.0
Out[25]: 2.0

-- 
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
  that is made terrible by our own mad attempt to interpret it as though it had
  an underlying truth."
   -- Umberto Eco