This code snippet provides methods to convert between various ieee754 floating point numbers format. For example double precision to single precision. The format is given by a pair (w, p) giving the size in bits of the exponent part and the significand part in the ieee754 representation of a number (these are (11, 52) for a standard 8 byte float and (8, 23) for a single precision float, but the snippet allows exotic sizes like (3, 4) or (9, 47)). Conversion to and from python floats are provided, given that your architecture's python floats have 8 bytes. Python >= 2.7 is required.

Edited by Gribouillis: documentation bugfix

#!/usr/bin/env python
# -*-coding: utf8-*-
# Title: anyfloat.py
# Author: Gribouillis for the python forum at www.daniweb.com
# Created: 2012-05-02 06:46:42.708131 (isoformat date)
# License: Public Domain
# Use this code freely.

"""Conversion module between floating point representations.

This module handles floating point values encoded with the IEEE 754 format. The following
is assumed about this format:

    * Real numbers are represented in 3 fields of bits:
        b bbbbbbbb bbbbbbbbbbbbbbbbbbbbbbb  (b = 0 or 1)
        + the first field is the _sign_ field (width = 1 bit)
        + the second field is the _exponent_ field (width = w bits)
        + the third field is the _significand_ field (width = p bits)
        Typical values are (w, p) = (8, 23) for single precision floating point numbers
        and (w, p) = (11, 52) for double precision floating point numbers. The number
        p is called the precision for this representation. The exponent and the significand
        of a real number are the integers represented by the exponent and significand fields
        in binary notation. For example, the number
        0 00000101 00000000000000000010000
        has exponent = 5 and significand = 16.
    * Apart from special values, there are two kind of numbers:

        - _normal_ numbers have an exponent between 00000001 and 11111110 in our example,
        more precisely: 1 <= exponent < 2^w - 1. They represent the real value
            val = +/- 1.significand x 2^(exponent - bias)
        The bias is 2^(w-1) - 1. The sign bit gives the +/-  part which is - if the sign bit is set.

        - _subnormal_ numbers have an exponent = 0, they represent the real value
            val = +/- 0.significand x 2^(1 - bias)
    * Special values have exponent = 2^w - 1 (11111111 in our example) they are
        - +/- infinity, which have significand = 0
        - NaN which have a non-zero significand and a zero sign bit.

More information about the IEEE floating point number format can be found at

This module offers a class "anyfloat" which is a named tuple with 3 fields: sign, exponent, significand,
but without restrictions on the size of the exponent and the significand.

An instance of anyfloat represents the real value

    val = +/- 0.significand x 2^(1 + exponent)

    * The sign field has value 1 for a negative value and 0 otherwise.
    * The exponent is the logarithm of |val| in base 2 (except for special values)
    * The significand is >= 0 except for special values:
        + significand = -1 is used to represent +/- infinity. In this case, exponent = 0
        + significand = -2 is used to represent NaN value. In this case, exponent = 0
Methods are provided to convert ieee numbers or python floats to/from anyfloat:

    anyfloat.to_ieee(size) converts to an integer which binary representation is the ieee754
        representation of this anyfloat with given size = (w, p).
    anyfloat.from_ieee(n, size) creates an anyfloat from an integer containing the ieee754
        representation of a value.
    anyfloat.__float__() converts to a python float.
    anyfloat.from_float(val) creates an anyfloat from a python floating point number.
Numbers can be converted from one ieee754 format to another, for example:

    >>> n = int("00000010100000000000000000010000", 2) # a single precision ieee number.
    >>> af = anyfloat.from_ieee(n), (8, 23))
    >>> n = af.to_ieee(af, (11, 52)) # convert to the double precision ieee number.

These conversions may occasionaly transform non zero values into +/-0 or +/- infinity
if the value is too small or too big for the target format.

from __future__ import print_function
from collections import namedtuple
from math import isnan
import struct
import sys

version_info = (1, 0)

if sys.version_info < (2, 7):
    raise ImportError("Module anyfloat requires python 2.7 or newer.")

DEFAULT_SIZE = (11, 52)

def trunc_round(n, k):
    rshift = n.bit_length() - 1 - k
    if rshift >= 0:
        n >>= (rshift)
        n <<= (-rshift)
    return (n + 1) >> 1
def more_bin_digits(n, k):
    return bool(n >> k)
def unset_high_bit(n):
    assert n > 0
    return n ^ (1 << (n.bit_length() - 1))
def fbin(n, nbits):
    assert (0 <= n)
    assert not (n >> nbits)
    return "{val:0>{width}}".format(val = bin(n)[2:], width = nbits)

_anyfloat = namedtuple("anyfloat", "sign exponent significand")

class anyfloat(_anyfloat):
    __slots__ = ()
    _b32 = 1 << 32
    _b64 = 1 << 64
    def __new__(cls, sign, exponent, significand):
        assert sign in (0, 1)
        if significand:
            significand = significand//(significand & -significand)
        return _anyfloat.__new__(cls, sign, exponent, significand)

    def _encode(log2, mantissa, a, b):
        A = ~(~0 << a)
        AA = A >> 1
        if mantissa <= 0:
            return ( (A, 0) if (mantissa == -1) else (A, 1 << (b-1)) ) if mantissa else (0, 0)
        elif log2 <= - AA:
            nbits = b + log2 + AA
            rounded = trunc_round(mantissa, nbits) if (nbits >= 0) else 0
            return (1, 0) if more_bin_digits(rounded, b) else (0, rounded)
        elif log2 <= AA:
            rounded = trunc_round(mantissa, b + 1)
            return (( (log2 + 1 + AA, 0) if (log2 < AA) else (A, 0) )
            if more_bin_digits(rounded, b+1) else (log2 + AA, unset_high_bit(rounded)) )
            return (A, 0)

    def _decode(exponent, significand, a, b):
        A = ~(~0 << a)
        AA = A >> 1
        assert 0 <= exponent <= A
        assert 0 <= significand < (1 << b)
        if exponent == A:
            return (0, -2 if significand else -1)
        elif exponent: # normal case
            return (exponent - AA, significand|(1 << b))
        else: # subnormal case
            if significand:
                return (significand.bit_length() - AA - b, significand)
                return (0, 0)

    def __float__(self):
        return self.int64_to_float(self.to_ieee())
    def from_float(cls, x):
        """Create an anyfloat instance from a python float (64 bits double precision number)."""
        return cls.from_ieee(cls.float_to_int64(x))
    def from_ieee(cls, n, size = DEFAULT_SIZE):
        """Create an anyfloat from an ieee754 integer.
        Create an anyfloat from an integer which binary representation is the ieee754
        format of a floating point number. The argument 'size' is a tuple (w, p)
        containing the width of the exponent part and the significand part in
        this ieee754 format."""
        w, p = size
        r = n >> p
        significand = (r << p) ^ n
        sign = int(r >> w)
        if not sign in (0, 1):
            raise ValueError(("Integer value out of range for ieee754 format", n, size))
        exponent = (sign << w) ^ r
        e, s = cls._decode(exponent, significand, w, p)
        if e == -2:
            sign = 0
        return cls(sign, e, s)

    def ieee_parts(self, size = DEFAULT_SIZE):
        w, p = size
        e, s = self._encode(self.exponent, self.significand, w, p)
        sign = 0 if (e + 1) >> w else self.sign
        return sign, e, s

    def to_ieee(self, size = DEFAULT_SIZE):
        """Convert to an ieee754 integer.
        Convert self to an integer which binary representation is the ieee754 format corresponding
        to the 'size' argument (read the documentation of from_ieee() for the meaning of the size
        sign, e, s = self.ieee_parts(size)
        return (((sign << size[0]) | e) << size[1]) | s
    def int64_to_float(cls, n):
        """Convert a 64 bits integer to a python float.
        This class method converts an integer representing a 64 bits floating point
        number in the ieee754 double precision format to this floating point number."""
        if not (0 <= n < cls._b64):
            raise ValueError(("Integer value out of range for 64 bits ieee754 format", n))
        u, v = divmod(n, cls._b32)
        return struct.unpack(">d", struct.pack(">LL", u, v))[0]
    def float_to_int64(cls, x):
        """Convert a python float to a 64 bits integer.
        This class method converts a float to an integer representing this
        float in the 64 bits ieee754 double precision format."""
        u, v = struct.unpack(">LL", struct.pack(">d", x))
        return (u << 32) | v
    def bin(self, size = DEFAULT_SIZE, sep=' '):
        """Return a binary representation of self.
        The returned string contains only the characters '0' and '1' and shows the
        ieee754 representation of the real number corresponding to self whith the given
        size = (w, p).
        if sep:
            sign, e, s = self.ieee_parts(size)
            return sep.join((fbin(sign, 1), fbin(e, size[0]), fbin(s, size[1])))
            return fbin(self.to_ieee(size), sum(size) + 1)

def main():
    from math import exp
    val = exp(2)
    print ("exp(2) =", val)
    af = anyfloat.from_float(val)
    print (af)
    print (af.bin(), "(64 bits float)")
    print (" " * 2, af.bin(size = (8,23)), "(32 bits)")
    print (" " * 7, af.bin(size = (3, 4)), "(8 bits)")
    print ("conversion to float works:", float(af) == val)
if __name__ == "__main__":
""" my output -->
exp(2) = 7.38905609893
anyfloat(sign=0, exponent=2, significand=4159668786720471)
0 10000000001 1101100011100110010010111000110101001101110110101110 (64 bits float)
   0 10000001 11011000111001100100110 (32 bits)
        0 101 1110 (8 bits)
conversion to float works: True

Notice that during the conversion to shorter formats, for example 64 bits to 32 bits,
the significand is not only truncated to 23 bits, it is also rounded to the nearest value
depending on the value of the next bit.
5 Years
Discussion Span
Last Post by pyTony

Thanks for your shared work, came to my mind that extraction of base 2 logarithm from floating point representation should be possible without using math log function. But maybe regular log function is currently so fast in hardware that it is not worth to special case this log2 case, or do you know something more about this? As from this it is only one division more to calculate logarithm in any base.


The module never uses math.log(). The logarithm of base 2 can indeed be extracted from the floating point representation, since this is exactly what this module does (the exponent part in anyfloat objects is the log with base 2). I prefer to extract the log from the representation, because the result is guaranteed: I don't want to deal with the issue of converting the result of math.log() to int. For subnormal numbers, the exponent part is not sufficient to get the log: the order of the significand's highest bit matters. However, it would certainly be faster to use math.log().

IMPORTANT NOTE: I found a typo at line 185. The literal number 8 must be replaced by the variable w.

Edited by Gribouillis


IMPORTANT NOTE: I found a typo at line 185. The literal number 8 must be replaced by the variable w.

I fixed that in your original code snippet.

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