问题描述

如何将字符串作为 32 位浮点数的二进制 IEEE 754 表示?

How to get the string as binary IEEE 754 representation of a 32 bit float?

示例

1.00 -> '00111111100000000000000000000000'

1.00 -> '00111111100000000000000000000000'

推荐答案

您可以使用 struct 包来做到这一点:

You can do that with the struct package:

import struct
def binary(num):
    return ''.join('{:0>8b}'.format(c) for c in struct.pack('!f', num))

将其打包为网络字节序浮点数，然后将每个生成的字节转换为 8 位二进制表示并将它们连接起来:

That packs it as a network byte-ordered float, and then converts each of the resulting bytes into an 8-bit binary representation and concatenates them out:

>>> binary(1)
'00111111100000000000000000000000'

编辑:有人要求扩大解释.我将使用中间变量对每个步骤进行注释.

Edit: There was a request to expand the explanation. I'll expand this using intermediate variables to comment each step.

def binary(num):
    # Struct can provide us with the float packed into bytes. The '!' ensures that
    # it's in network byte order (big-endian) and the 'f' says that it should be
    # packed as a float. Alternatively, for double-precision, you could use 'd'.
    packed = struct.pack('!f', num)
    print 'Packed: %s' % repr(packed)

    # For each character in the returned string, we'll turn it into its corresponding
    # integer code point
    # 
    # [62, 163, 215, 10] = [ord(c) for c in '>xa3xd7
']
    integers = [ord(c) for c in packed]
    print 'Integers: %s' % integers

    # For each integer, we'll convert it to its binary representation.
    binaries = [bin(i) for i in integers]
    print 'Binaries: %s' % binaries

    # Now strip off the '0b' from each of these
    stripped_binaries = [s.replace('0b', '') for s in binaries]
    print 'Stripped: %s' % stripped_binaries

    # Pad each byte's binary representation's with 0's to make sure it has all 8 bits:
    #
    # ['00111110', '10100011', '11010111', '00001010']
    padded = [s.rjust(8, '0') for s in stripped_binaries]
    print 'Padded: %s' % padded

    # At this point, we have each of the bytes for the network byte ordered float
    # in an array as binary strings. Now we just concatenate them to get the total
    # representation of the float:
    return ''.join(padded)

还有几个例子的结果:

>>> binary(1)
Packed: '?x80x00x00'
Integers: [63, 128, 0, 0]
Binaries: ['0b111111', '0b10000000', '0b0', '0b0']
Stripped: ['111111', '10000000', '0', '0']
Padded: ['00111111', '10000000', '00000000', '00000000']
'00111111100000000000000000000000'

>>> binary(0.32)
Packed: '>xa3xd7
'
Integers: [62, 163, 215, 10]
Binaries: ['0b111110', '0b10100011', '0b11010111', '0b1010']
Stripped: ['111110', '10100011', '11010111', '1010']
Padded: ['00111110', '10100011', '11010111', '00001010']
'00111110101000111101011100001010'

这篇关于Python中浮点数的二进制表示(位不是十六进制)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！