TP <tribulati...@paralleles.invalid> wrote:
> Hi everybody,
>
> Try the following python statements:
>
> >>> "%.40f" % 0.2222222222222222222222222222222
> '0.2222222222222222098864108374982606619596'
> >>> float( 0.2222222222222222222222222222222)
> 0.22222222222222221
>
> It seems the first result is the same than the following C program:
> ################
> #include <stdio.h>
>
> int main(void)
> {
> double a = 0.2222222222222222222222222222222;
>
> printf( "%.40f\n", a );
> return 0;
> }
> #################
>
> My problem is the following:
> * the precision "40" (for example) is given by the user, not by the
> programmer.
> * I want to use the string conversion facility with specifier "e", that
> yields number is scientific format; so I cannot apply float() on the result
> of "%.40e" % 0.2222222222222222222222222222222, I would lost the scientific
> format.
>
> Is there any means to obtain the full C double in Python, or should I limit
> the precision given by the user (and used in "%.*e") to the one of a Python
> float?
Python floats are actually C doubles (as you proved yourself with your
little test program).
Eg
>>> 1.+2.**-52
1.0000000000000002
>>> 1.+2.**-53
1.0
Indicating that python floats have about 52 bits of precision, so are
definitely what C calls doubles.
When you do
>>> float( 0.2222222222222222222222222222222)
0.22222222222222221
Python prints as many decimal places as are significant in the answer.
This is covered in the FAQ
http://www.python.org/doc/faq/general/#why-are-floating-point-calculations-so-inaccurate
If you want more precision use the built in decimal module or the
third party gmpy module.
--
Nick Craig-Wood <n...@craig-wood.com> -- http://www.craig-wood.com/nick
--
http://mail.python.org/mailman/listinfo/python-list
