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author | Raymond Hettinger <python@rcn.com> | 2009-06-28 23:21:38 (GMT) |
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committer | Raymond Hettinger <python@rcn.com> | 2009-06-28 23:21:38 (GMT) |
commit | 1d1806843b7756803b20d504126433b775198b6b (patch) | |
tree | 139ae35ad5b555e4636221a843aee07df7cd2d8e /Doc/tutorial/floatingpoint.rst | |
parent | e9eb7b6581bd0a9430a117efcfb1c65658fb4cd6 (diff) | |
download | cpython-1d1806843b7756803b20d504126433b775198b6b.zip cpython-1d1806843b7756803b20d504126433b775198b6b.tar.gz cpython-1d1806843b7756803b20d504126433b775198b6b.tar.bz2 |
Clean-up floating point tutorial.
Diffstat (limited to 'Doc/tutorial/floatingpoint.rst')
-rw-r--r-- | Doc/tutorial/floatingpoint.rst | 51 |
1 files changed, 28 insertions, 23 deletions
diff --git a/Doc/tutorial/floatingpoint.rst b/Doc/tutorial/floatingpoint.rst index 2db1842..0230183 100644 --- a/Doc/tutorial/floatingpoint.rst +++ b/Doc/tutorial/floatingpoint.rst @@ -50,7 +50,7 @@ decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base Stop at any finite number of bits, and you get an approximation. On most machines today, floats are approximated using a binary fraction with -the numerator using the first 53 bits following the most significant bit and +the numerator using the first 53 bits starting with the most significant bit and with the denominator as a power of two. In the case of 1/10, the binary fraction is ``3602879701896397 / 2 ** 55`` which is close to but not exactly equal to the true value of 1/10. @@ -230,12 +230,8 @@ as :: and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``), the best value for *N* is 56:: - >>> 2**52 - 4503599627370496 - >>> 2**53 - 9007199254740992 - >>> 2**56/10 - 7205759403792794.0 + >>> 2**52 <= 2**56 // 10 < 2**53 + True That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits. The best possible value for *J* is then that quotient rounded:: @@ -250,14 +246,13 @@ by rounding up:: >>> q+1 7205759403792794 -Therefore the best possible approximation to 1/10 in 754 double precision is -that over 2\*\*56, or :: +Therefore the best possible approximation to 1/10 in 754 double precision is:: - 7205759403792794 / 72057594037927936 + 7205759403792794 / 2 ** 56 Dividing both the numerator and denominator by two reduces the fraction to:: - 3602879701896397 / 36028797018963968 + 3602879701896397 / 2 ** 55 Note that since we rounded up, this is actually a little bit larger than 1/10; if we had not rounded up, the quotient would have been a little bit smaller than @@ -269,24 +264,34 @@ above, the best 754 double approximation it can get:: >>> 0.1 * 2 ** 55 3602879701896397.0 -If we multiply that fraction by 10\*\*60, we can see the value of out to -60 decimal digits:: +If we multiply that fraction by 10\*\*55, we can see the value out to +55 decimal digits:: - >>> 3602879701896397 * 10 ** 60 // 2 ** 55 + >>> 3602879701896397 * 10 ** 55 // 2 ** 55 1000000000000000055511151231257827021181583404541015625 -meaning that the exact number stored in the computer is approximately equal to -the decimal value 0.100000000000000005551115123125. Rounding that to 17 -significant digits gives the 0.10000000000000001 that Python displays (well, -will display on any 754-conforming platform that does best-possible input and -output conversions in its C library --- yours may not!). +meaning that the exact number stored in the computer is equal to +the decimal value 0.1000000000000000055511151231257827021181583404541015625. +Instead of displaying the full decimal value, many languages (including +older versions of Python), round the result to 17 significant digits:: + + >>> format(0.1, '.17f') + '0.10000000000000001' The :mod:`fractions` and :mod:`decimal` modules make these calculations easy:: >>> from decimal import Decimal >>> from fractions import Fraction - >>> print(Fraction.from_float(0.1)) - 3602879701896397/36028797018963968 - >>> print(Decimal.from_float(0.1)) - 0.1000000000000000055511151231257827021181583404541015625 + + >>> Fraction.from_float(0.1) + Fraction(3602879701896397, 36028797018963968) + + >>> (0.1).as_integer_ratio() + (3602879701896397, 36028797018963968) + + >>> Decimal.from_float(0.1) + Decimal('0.1000000000000000055511151231257827021181583404541015625') + + >>> format(Decimal.from_float(0.1), '.17') + '0.10000000000000001' |