B. Floating Point Arithmetic: Issues and Limitations

Floating-point numbers are represented in computer hardware as base 2 (binary) fractions. For example, the decimal fraction

0.125

has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction

0.001

has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only real difference being that the first is written in base 10 fractional notation, and the second in base 2.

Unfortunately, most decimal fractions cannot be represented exactly as binary fractions. A consequence is that, in general, the decimal floating-point numbers you enter are only approximated by the binary floating-point numbers actually stored in the machine.

The problem is easier to understand at first in base 10. Consider the fraction 1/3. You can approximate that as a base 10 fraction:

0.3

or, better,

0.33

or, better,

0.333

and so on. No matter how many digits you're willing to write down, the result will never be exactly 1/3, but will be an increasingly better approximation of 1/3.

In the same way, no matter how many base 2 digits you're willing to use, the decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base 2, 1/10 is the infinitely repeating fraction

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 (implicitly) uses the builtin
`repr()` function to obtain a string version of everything it
displays. For floats, `repr(`

rounds the true
decimal value to 17 significant digits, giving
`float`)

0.10000000000000001

`repr(`

produces 17 significant digits because it
turns out that's enough (on most machines) so that
`float`)`eval(repr(`

exactly for all finite floats
`x`)) == `x``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(`

to reproduce `x`))`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.

Another consequence is that since 0.1 is not exactly 1/10, adding 0.1 to itself 10 times may not yield exactly 1.0, either:

>>> sum = 0.0 >>> for i in range(10): ... sum += 0.1 ... >>> sum 0.99999999999999989

Binary floating-point arithmetic holds many surprises like this. The
problem with "0.1" is explained in precise detail below, in the
"Representation Error" section. See
*The Perils of Floating
Point* for a more complete account of other common surprises.

As that says near the end, ``there are no easy answers.'' Still, don't be unduly wary of floating-point! The errors in Python float operations are inherited from the floating-point hardware, and on most machines are on the order of no more than 1 part in 2**53 per operation. That's more than adequate for most tasks, but you do need to keep in mind that it's not decimal arithmetic, and that every float operation can suffer a new rounding error.

While pathological cases do exist, for most casual use of
floating-point arithmetic you'll see the result you expect in the end
if you simply round the display of your final results to the number of
decimal digits you expect. `str()` usually suffices, and for
finer control see the discussion of Python's `%`

format
operator: the `%g`

, `%f`

and `%e`

format codes
supply flexible and easy ways to round float results for display.

B.1 Representation Error

This section explains the ``0.1'' example in detail, and shows how you can perform an exact analysis of cases like this yourself. Basic familiarity with binary floating-point representation is assumed.

*Representation error* refers to that some (most, actually)
decimal fractions cannot be represented exactly as binary (base 2)
fractions. This is the chief reason why Python (or Perl, C, C++,
Java, Fortran, and many others) often won't display the exact decimal
number you expect:

>>> 0.1 0.10000000000000001

Why is that? 1/10 is not exactly representable as a binary fraction.
Almost all machines today (November 2000) use IEEE-754 floating point
arithmetic, and almost all platforms map Python floats to IEEE-754
"double precision". 754 doubles contain 53 bits of precision, so on
input the computer strives to convert 0.1 to the closest fraction it can
of the form `J`/2**`N` where `J` is an integer containing
exactly 53 bits. Rewriting

1 / 10 ~= J / (2**N)

as

J ~= 2**N / 10

and recalling that `J` has exactly 53 bits (is `>= 2**52`

but
`< 2**53`

), the best value for `N` is 56:

>>> 2**52 4503599627370496L >>> 2L**53 9007199254740992L >>> 2L**56/10 7205759403792793L

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:

>>> q, r = divmod(2L**56, 10) >>> r 6L

Since the remainder is more than half of 10, the best approximation is obtained by rounding up:

>>> q+1 7205759403792794L

Therefore the best possible approximation to 1/10 in 754 double precision is that over 2**56, or

7205759403792794 / 72057594037927936

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 1/10. But in no case can it be *exactly* 1/10!

So the computer never ``sees'' 1/10: what it sees is the exact fraction given above, the best 754 double approximation it can get:

>>> .1 * 2L**56 7205759403792794.0

If we multiply that fraction by 10**30, we can see the (truncated) value of its 30 most significant decimal digits:

>>> 7205759403792794L * 10L**30 / 2L**56 100000000000000005551115123125L

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!).

See