When is divmod(a,b)[0] == floor(a/b)-1 ?
Robert Kern
robert.kern at gmail.com
Thu Sep 24 22:55:00 CEST 2009
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
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