Tutorial Python

0.0001100110011001100110011001100110011001100110011...
Stop at any finite number of bits, and you get an approximation. This is why you see things like:
>>> 0.1
0.10000000000000001
On most machines today, that is what you’ll see if you enter 0.1 at a Python prompt. You may not, though, because
the number of bits used by the hardware to store floating-point values can vary across machines, and Python only
prints a decimal approximation to the true decimal value of the binary approximation stored by the machine. On
most machines, if Python were to print the true decimal value of the binary approximation stored for 0.1, it would
have to display
>>> 0.1
0.1000000000000000055511151231257827021181583404541015625
instead! The Python prompt uses the builtin repr() function to obtain a string version of everything it displays.
For floats, repr(float) rounds the true decimal value to 17 significant digits, giving
0.10000000000000001
repr(float) produces 17 significant digits because it turns out that’s enough (on most machines) so that
eval(repr(x)) == x exactly for all finite floats x, but rounding to 16 digits is not enough to make that
true.
Note that this is in the very nature of binary floating-point: this is not a bug in Python, it is not a bug in your code
either. You’ll see the same kind of thing in all languages that support your hardware’s floating-point arithmetic
(although some languages may not display the difference by default, or in all output modes).
Python’s builtin str() function produces only 12 significant digits, and you may wish to use that instead. It’s
unusual for eval(str(x)) to reproduce x, but the output may be more pleasant to look at:
>>> print str(0.1)
0.1
It’s important to realize that this is, in a real sense, an illusion: the value in the machine is not exactly 1/10, you’re
simply rounding the display of the true machine value.
Other surprises follow from this one. For example, after seeing
>>> 0.1
0.10000000000000001
you may be tempted to use the round() function to chop it back to the single digit you expect. But that makes
no difference:
>>> round(0.1, 1)
0.10000000000000001
The problem is that the binary floating-point value stored for "0.1"was already the best possible binary approximation
to 1/10, so trying to round it again can’t make it better: it was already as good as it gets.
92 Apêndice B. Floating Point Arithmetic: Issues and Limitations