Everybody knows that "floating point numbers cause problems." Many people have misconceptions about what those problems are, and how to deal with them. There are some great references for people who want to understand things, the (justly) most famous being What Every Computer Scientist Should Know About Floating-Point Arithmetic by David Goldberg.

But many Python programmers like to learn by experimenting. That's why you've chosen a language with an interactive REPL, powerful introspection, and, when it comes down to it, a readable open-source implementation, right?

So, let's look at how floating point numbers are represented in Python; understanding what all the bits mean may make it easier to understand how rounding errors and so forth work.

On nearly any platform you care about, floating point numbers—both binary (float) and decimal (Decimal)—are going to be stored in IEEE 754-2008 formats. So, let's look at how those work.

IEEE binary floats

Scientific notation

IEEE floats are based on scientific notation, but in binary.

Scientific notation looks like:

-1.042e3

The "-" sign is obvious. The "1.042" is almost obvious, but note that in "normalized form" there always must be exactly one nonzero digit left of the "." (that's what makes it scientific notation), and you're allowed to have trailing zeros (if, e.g., you had 11 to 4 significant digits, you'd write 1.100e1, not 1.1e1). The "e3" means that you multiple by 10**3.

So:

- (1 + 0/10 + 4/100 + 2/1000) * 10**3 == -104.2

For binary scientific notation, everything works the same way, except with 2 in place of 10. So:

-1.01e2

- (1 + 0/2 + 1/4) * 2**2 = -101 (binary) or -5 (decimal) 

Representation in bits

Obviously, any float based on scientific notation is going to have three things to store: the sign, the mantissa (the decimal 1.042 or binary 1.01), and the exponent (the 3 in 10**3 or 2 in 2**2).

• 1 sign bit
• w exponent bits
• p-1 significand bits

Let's pick a specific, very small, size to make our lives easy: just 3 exponent bits and 2 significand bits. Of course in real life you'd never deal with 6-bit floats, but having only 64 possible values means we can look at an exhaustive table of them (which I've included below), which can be very helpful.

We'll call this 6-bit format "binary6", even though technically it isn't a legal interchange format. (For one thing, interchange formats must have either 16 bits, or a multiple of 32 other than 96. For another, a k-bit interchange format for any k but 16 and 32 must have w = round(4 * math.log2(k)) − 13, which in our case would give -3 bits.)

For the binary64 format, everything is exactly the same as our binary6 format, except that w is 11 instead of 3, and p-1 is 52 instead of 2.

Sign

The sign is easy. Unlike integers, there's no 2s complement or anything fancy going on here. You compute the number from the exponent and significand, and then if the sign bit is 1, you negate it. So, -5.0 looks exactly like 5.0 except for the first bit. This does have some strange consequences—it means -0.0 and +0.0 are different numbers, and it means the bits for the smallest negative number and the smallest positive number are very far apart—but it also makes a lot of things easier. So, in our example of -1.01e2, the sign bit is 1.

Mantissa and significand

Next, we've got the mantissa, the 1.01 part. But notice that we don't have 3 mantissa bits, we have 2 significand bits. How does this work? Well, in scientific notation, there's always exactly one non-zero digit to the left of the decimal point. In binary, the only non-zero digit is 1, so the format saves a bit by not storing that 1.

So, for 2 bits, the possible values are 1.00 (stored as 00), 1.01 (as 01), 1.10 (as 10), and 1.11 (as 11).

Exponent

Finally, we've got the exponent of 2. To handle the possibility of negative exponents, instead of storing the exponent itself, we store the biased exponent—that is, we add 2**(w-1)-1. Also, the smallest and largest values are reserved for special purposes, which we'll get to in a moment. For 3 bits, the possible values are special (stored as 000), -2 (as 001), -1 (as 010), 0 (as 011), 1 (as 100), 2 (as 101), 3 (as 110), and special (as 111).

So, our number, -1.01e2, is stored as 1 (sign) 101 (exponent) 01 (significand), or 110101.

Infinity, NaN, and subnormals

The smallest exponent value is used to store 0. After all, if there were no special rule, 0 000 00 would mean +1.00 * 2**-3, which is obviously not 0. But just saying exponent 0 means the value is 0 is a waste of bits—that would mean we have 8 different zero values (0 000 00, 0 000 01, 0 000 10, 0 000 11, 1 000 00, 1 000 01, 1 000 10, and 1 000 11). Having positive and negative zero can sometimes be useful, but having 4 different positive zeros is pointless.

So, instead, a 000 exponent actually means the same thing as a 001 exponent, except that a 0 instead of a 1 is prepended to the significand to get the mantissa. A number like 0.01e-2 isn't properly normalized scientific notation—that would be 1.00e-4. So, these numbers are called "subnormal" or "denormal".

Meanwhile, the largest exponent value is used to store infinity. Again, though, that would give us 8 infinities when we only need 2. So, if the significand is nonzero, instead of infinity, it's NaN. The first bit of the significand is the quiet bit, and the remaining bits (in this case, just 1) are the payload, which is reserved for anything a user-level library wants.

Table

So, in our 6-bit format, here's what all the possible numbers mean:

    sign exponent significand    binary scientific  decimal
       0      000          00    0.00e-2 = 0        0.0
       0      000          01    0.01e-2 = 1e-4     0.0625
       0      000          10    0.10e-2 = 1e-3     0.125
       0      000          11    0.11e-2 = 1.1e-3   0.1875
       0      001          00    1.00e-2            0.25
       0      001          01    1.01e-2            0.3125
       0      001          10    1.10e-2            0.375
       0      001          11    1.11e-2            0.4375
       0      010          00    1.00e-1            0.5
       0      010          01    1.01e-1            0.625
       0      010          10    1.10e-1            0.75
       0      010          11    1.11e-1            0.875
       0      011          00    1.00e0             1.
       0      011          01    1.01e0             1.25
       0      011          10    1.10e0             1.5
       0      011          11    1.11e0             1.75
       0      100          00    1.00e1             2.
       0      100          01    1.01e1             2.5
       0      100          10    1.10e1             3.
       0      100          11    1.11e1             3.5
       0      101          00    1.00e2             4.
       0      101          01    1.01e2             5.
       0      101          10    1.10e2             6.
       0      101          11    1.11e2             7
       0      110          00    1.00e3             8.
       0      110          01    1.01e3             10.
       0      110          10    1.10e3             12.
       0      110          11    1.11e3             14.
       0      111          00    inf                inf
       0      111          01    snan1              snan1
       0      111          10    qnan0              qnan0
       0      111          11    qnan1              qnan1
       1      000          00    -0.00e-2 = -0      -0.0
       1      000          01    -0.01e-2 = -1e-4   -0.0625
       1      000          10    -0.10e-2 = -1e-3   -0.125
       1      000          11    -0.11e-2 = -1.1e-3 -0.1875
       1      001          00    -1.00e-2           -0.25
       1      001          01    -1.01e-2           -0.3125
       1      001          10    -1.10e-2           -0.375
       1      001          11    -1.11e-2           -0.4375
       1      010          00    -1.00e-1           -0.5
     ...
       1      110          00    1.00e3             -8.
       1      110          01    1.01e3             -10.
       1      110          10    1.10e3             -12.
       1      110          11    1.11e3             -14.
       1      111          00    -inf               -inf
       1      111          01    -snan1             -snan1
       1      111          10    -qnan0             -qnan0
       1      111          11    -qnan1             -qnan1

Rounding

Given the table above, rounding errors should be pretty easy to understand.

If you want to store 1.1 as a binary6 float, the closest value is 1.0 (0 100 00), which is off by 0.1; the next best would be 1.25 (0 100 01), which is off by even more 0.15.

Now, let's say you want to test 0.6 / 4 == 0.15. The closest binary6 values to start with are 0.625 and 4. Divide those, and the closest result is 0.1875. But the closest value to 0.15 is 0.125, and 0.1875 != 0.125.

You can work out the same thing with more familiar examples with binary64, like 0.3 / 3 == 0.1.

Reading floats as ints

Manipulating float bits in Python

OK, now we know how a float is stored. But how can we play with this in Python?

While it's possible (at least in CPython) to get at the values under the covers via ctypes, it's generally simpler (and a lot safer) to just use the struct module to pack a float in big-endian C double format. Then we've got the bits as a sequence of bytes—or, if you prefer, you can turn it into a 64-bit unsigned integer, or use a third-party library like bitarray or bitstring to make your life a whole lot easier.

So, let's look at how -123.456 is stored as a float:

    >>> f = -123.456
    >>> b = bitstring.pack('>d', f)
    >>> sbit, wbits, pbits = b[:1], b[1:12], b[12:]
    >>> sbit.bin
    '1'

That's easy: sign bit 1 means negative.

    >>> wbits.bin
    '10000000101'
    >>> wbits.uint - (1<<10) + 1
    6

So, the exponent is (decimal) 6.

    >>> pbits.bin
    '1110110111010010111100011010100111111011111001110111'
    >>> 1 + pbits.uint / (1<<52)
    1.929

So, the mantissa is (decimal) 1.929.

So, the number is -1.929 * 2**6. Which checks out:

    >>> -1.929 * 2**6
    -123.456

Or, if you prefer:

    >>> pbits.hex
    'edd2f1a9fbe77'

So, in Python's float hex format, the number is -1.edd2f1a9fbe77p+6, which we can check:

    >>> f.hex()
    '-0x1.edd2f1a9fbe77p+6'

You can also use this to create and parse NaN payloads, which Python doesn't do natively:

    >>> quiet, payload = 1, 3
    >>> bits = (0b11111111111 << 52) | (quiet << 51) | payload
    >>> qnan3 = struct.unpack('>d', struct.pack('>Q', bits))[0]

Of course Python is going to print qnan3 as just plain "nan", but if your platform has some way of printing out non-default NaN values, you may be able to see it like this:

    >>> ctypes.printf(b'%f', ctypes.c_double(qnan3))
    QNaN3

And if you want to do the nextafter or float_difference hacks described in the last section:

    >>> f = 1.1
    >>> d = struct.unpack('>Q', struct.pack('>d', f))[0]
    >>> struct.unpack('>d', struct.pack('>Q', d-1))[0]
    1.0999999999999999
    >>> struct.unpack('>d', struct.pack('>Q', d+1))[0]
    1.1000000000000003

The floatextras package

Now that you know how to do this all manually, the floatextras package on PyPI wraps it all up for you:

    >>> f = 1.1
    >>> floatextras.next_plus(f)
    1.1000000000000003
    >>> floatextras.float_difference(1.0999999999999999, 1.1000000000000003)
    -2
    >>> floatextras.as_tuple(-123.456)
    FloatTuple(sign=1, digits=(1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1), exponent=6)
    >>> floatextras.as_tuple(qnan3)
    FloatTuple(sign=0, digits=(1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1), exponent=1024)

Caveat for some platforms

Decimal floats