finding out the precision of floats

[Arnaud Delobelle]

On Feb 25, 1:31 pm, "John Machin" <sjmac... at lexicon.net> wrote:
> On Feb 25, 11:06 pm, "Arnaud Delobelle" <arno... at googlemail.com>
> wrote:
[...]
> Evidently not; here's some documentation we both need(ed) to read:
>
> http://docs.python.org/tut/node16.html

Thanks for this link

> I'm very curious to know what the exceptions were in November 2000 and
> if they still exist. There is also the question of how much it matters
> to you. Presuming the representation is 64 bits, even taking 3 bits
> off the mantissa and donating them to the exponent leaves you with
> 15.05 decimal digits -- perhaps you could assume that you've got at
> least 15 decimal digits.

It matters to me because I prefer to avoid making assumptions in my
code!
Moreover the reason I got interested in this is because I am creating
a
Dec class (decimal numbers) and I would like that:
   * given a float x, float(Dec(x)) == x for as many values of x as
possible
   * given a decimal d, Dec(float(d)) == d for as many values of d as
possible.

(and I don't want the standard Decimal class :)

> While we're waiting for the gurus to answer, here's a routine that's
> slightly less dodgy than yours:
>
> | >>> for n in range(200):
> | ...     if (1.0 + 1.0/2**n) == 1.0:
> | ...         print n, "bits"
> | ...         break
> | ...
> | 53 bits

Yes I have a similar routine:

|def fptest():
|    "Gradually fill the mantissa of a float with 1s until running out
of bits"
|    ix = 0
|    fx = 0.0
|    for i in range(200):
|        fx = 2*fx+1
|        ix = 2*ix+1
|        if ix != fx:
|            return i

...

>>> fptest()
53

I guess it would be OK to use this.  It's just that this 'poking into
floats'
seems a bit strange to me ;)

Thanks for your help

--
Arnaud