#### /Doc/tutorial/floatingpoint.rst

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```  1.. _tut-fp-issues:
2
3**************************************************
4Floating Point Arithmetic:  Issues and Limitations
5**************************************************
6
7.. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net>
8
9
10Floating-point numbers are represented in computer hardware as base 2 (binary)
11fractions.  For example, the decimal fraction ::
12
13   0.125
14
15has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction ::
16
17   0.001
18
19has value 0/2 + 0/4 + 1/8.  These two fractions have identical values, the only
20real difference being that the first is written in base 10 fractional notation,
21and the second in base 2.
22
23Unfortunately, most decimal fractions cannot be represented exactly as binary
24fractions.  A consequence is that, in general, the decimal floating-point
25numbers you enter are only approximated by the binary floating-point numbers
26actually stored in the machine.
27
28The problem is easier to understand at first in base 10.  Consider the fraction
291/3.  You can approximate that as a base 10 fraction::
30
31   0.3
32
33or, better, ::
34
35   0.33
36
37or, better, ::
38
39   0.333
40
41and so on.  No matter how many digits you're willing to write down, the result
42will never be exactly 1/3, but will be an increasingly better approximation of
431/3.
44
45In the same way, no matter how many base 2 digits you're willing to use, the
46decimal value 0.1 cannot be represented exactly as a base 2 fraction.  In base
472, 1/10 is the infinitely repeating fraction ::
48
49   0.0001100110011001100110011001100110011001100110011...
50
51Stop at any finite number of bits, and you get an approximation.  This is why
52you see things like::
53
54   >>> 0.1
55   0.10000000000000001
56
57On most machines today, that is what you'll see if you enter 0.1 at a Python
58prompt.  You may not, though, because the number of bits used by the hardware to
59store floating-point values can vary across machines, and Python only prints a
60decimal approximation to the true decimal value of the binary approximation
61stored by the machine.  On most machines, if Python were to print the true
62decimal value of the binary approximation stored for 0.1, it would have to
63display ::
64
65   >>> 0.1
66   0.1000000000000000055511151231257827021181583404541015625
67
68instead!  The Python prompt uses the built-in :func:`repr` function to obtain a
69string version of everything it displays.  For floats, ``repr(float)`` rounds
70the true decimal value to 17 significant digits, giving ::
71
72   0.10000000000000001
73
74``repr(float)`` produces 17 significant digits because it turns out that's
75enough (on most machines) so that ``eval(repr(x)) == x`` exactly for all finite
76floats *x*, but rounding to 16 digits is not enough to make that true.
77
78Note that this is in the very nature of binary floating-point: this is not a bug
79in Python, and it is not a bug in your code either.  You'll see the same kind of
80thing in all languages that support your hardware's floating-point arithmetic
81(although some languages may not *display* the difference by default, or in all
82output modes).
83
84Python's built-in :func:`str` function produces only 12 significant digits, and
85you may wish to use that instead.  It's unusual for ``eval(str(x))`` to
86reproduce *x*, but the output may be more pleasant to look at::
87
88   >>> print str(0.1)
89   0.1
90
91It's important to realize that this is, in a real sense, an illusion: the value
92in the machine is not exactly 1/10, you're simply rounding the *display* of the
93true machine value.
94
95Other surprises follow from this one.  For example, after seeing ::
96
97   >>> 0.1
98   0.10000000000000001
99
100you may be tempted to use the :func:`round` function to chop it back to the
101single digit you expect.  But that makes no difference::
102
103   >>> round(0.1, 1)
104   0.10000000000000001
105
106The problem is that the binary floating-point value stored for "0.1" was already
107the best possible binary approximation to 1/10, so trying to round it again
108can't make it better:  it was already as good as it gets.
109
110Another consequence is that since 0.1 is not exactly 1/10, summing ten values of
1110.1 may not yield exactly 1.0, either::
112
113   >>> sum = 0.0
114   >>> for i in range(10):
115   ...     sum += 0.1
116   ...
117   >>> sum
118   0.99999999999999989
119
120Binary floating-point arithmetic holds many surprises like this.  The problem
121with "0.1" is explained in precise detail below, in the "Representation Error"
122section.  See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
123for a more complete account of other common surprises.
124
125As that says near the end, "there are no easy answers."  Still, don't be unduly
126wary of floating-point!  The errors in Python float operations are inherited
127from the floating-point hardware, and on most machines are on the order of no
128more than 1 part in 2\*\*53 per operation.  That's more than adequate for most
129tasks, but you do need to keep in mind that it's not decimal arithmetic, and
130that every float operation can suffer a new rounding error.
131
132While pathological cases do exist, for most casual use of floating-point
133arithmetic you'll see the result you expect in the end if you simply round the
134display of your final results to the number of decimal digits you expect.
135:func:`str` usually suffices, and for finer control see the :meth:`str.format`
136method's format specifiers in :ref:`formatstrings`.
137
138
139.. _tut-fp-error:
140
141Representation Error
142====================
143
144This section explains the "0.1" example in detail, and shows how you can perform
145an exact analysis of cases like this yourself.  Basic familiarity with binary
146floating-point representation is assumed.
147
148:dfn:`Representation error` refers to the fact that some (most, actually)
149decimal fractions cannot be represented exactly as binary (base 2) fractions.
150This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
151others) often won't display the exact decimal number you expect::
152
153   >>> 0.1
154   0.10000000000000001
155
156Why is that?  1/10 is not exactly representable as a binary fraction. Almost all
157machines today (November 2000) use IEEE-754 floating point arithmetic, and
158almost all platforms map Python floats to IEEE-754 "double precision".  754
159doubles contain 53 bits of precision, so on input the computer strives to
160convert 0.1 to the closest fraction it can of the form *J*/2\*\**N* where *J* is
161an integer containing exactly 53 bits.  Rewriting ::
162
163   1 / 10 ~= J / (2**N)
164
165as ::
166
167   J ~= 2**N / 10
168
169and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
170the best value for *N* is 56::
171
172   >>> 2**52
173   4503599627370496L
174   >>> 2**53
175   9007199254740992L
176   >>> 2**56/10
177   7205759403792793L
178
179That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits.  The
180best possible value for *J* is then that quotient rounded::
181
182   >>> q, r = divmod(2**56, 10)
183   >>> r
184   6L
185
186Since the remainder is more than half of 10, the best approximation is obtained
187by rounding up::
188
189   >>> q+1
190   7205759403792794L
191
192Therefore the best possible approximation to 1/10 in 754 double precision is
193that over 2\*\*56, or ::
194
195   7205759403792794 / 72057594037927936
196
197Note that since we rounded up, this is actually a little bit larger than 1/10;
198if we had not rounded up, the quotient would have been a little bit smaller than
1991/10.  But in no case can it be *exactly* 1/10!
200
201So the computer never "sees" 1/10:  what it sees is the exact fraction given
202above, the best 754 double approximation it can get::
203
204   >>> .1 * 2**56
205   7205759403792794.0
206
207If we multiply that fraction by 10\*\*30, we can see the (truncated) value of
208its 30 most significant decimal digits::
209
210   >>> 7205759403792794 * 10**30 / 2**56
211   100000000000000005551115123125L
212
213meaning that the exact number stored in the computer is approximately equal to
214the decimal value 0.100000000000000005551115123125.  Rounding that to 17
215significant digits gives the 0.10000000000000001 that Python displays (well,
216will display on any 754-conforming platform that does best-possible input and
217output conversions in its C library --- yours may not!).
218
219
```