Picking apart a floating point number

Copyright (C) 2020 Andreas Kloeckner

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In [1]:
import numpy as np
In [31]:
# Never mind the details of this function...



def left_pad(s, desired_len, pad="0"):

    if len(s) < desired_len:

        s = (desired_len - len(s)) * pad + s

    return s

        

def pretty_print_fp(x):

    print("---------------------------------------------")

    print("Floating point structure for %r" % x)

    print("---------------------------------------------")

    import struct

    s = struct.pack("d", x)



    def get_bit(i):

        byte_nr, bit_nr = divmod(i, 8)

        return int(bool(

            s[byte_nr] & (1 << bit_nr)

            ))



    def get_bits(lsb, count):

        return sum(get_bit(i+lsb)*2**i for i in range(count))



    # https://en.wikipedia.org/wiki/Double_precision_floating-point_format



    print("Sign bit (1:negative):", get_bit(63))

    exponent = get_bits(52, 11)

    print("Stored exponent bits: %d" % exponent)

    print("Exponent (with offset): %d" % (exponent - 1023))

    if exponent == 0:

        print("  In subnormal range. Effective exponent: -1022")

    fraction = get_bits(0, 52)

    print(f"Stored fraction bits: ({left_pad(bin(fraction)[2:], 52)}_2")

    if exponent != 0:

        significand = fraction + 2**52

    else:

        significand = fraction

    sig_str = left_pad(bin(significand)[2:], 53)

    print(f"Significand: ({sig_str[0]}.{sig_str[1:]})_2")

    print("Significand:", repr(significand / (2**52)))
In [32]:
pretty_print_fp(2**(-1025))

---------------------------------------------
Floating point structure for 2.781342323134e-309
---------------------------------------------
Sign bit (1:negative): 0
Stored exponent bits: 0
Exponent (with offset): -1023
In subnormal range. Effective exponent: -1022
Stored fraction bits: (0010000000000000000000000000000000000000000000000000_2
Significand: (0.0010000000000000000000000000000000000000000000000000)_2
Significand: 0.125

Things to try:

  • Twiddle the sign bit

  • 1,2,4,8

  • 0.5,0.25

  • $2^{\pm 1023}$, $2^{\pm 1024}$

  • float("nan")

In [ ]: