[英]Hexadecimal string to two 18-bit integer
I am new to hexadecimals and couldn't find any similar previous posts.我是十六进制的新手,找不到任何类似的以前的帖子。 I have a hexadecimal string
50f116bcf
, b3b4d25d0
, for example, that is known to be a 36-bit integer.例如,我有一个十六进制字符串
50f116bcf
, b3b4d25d0
,已知它是一个 36 位整数。 Not sure if it's signed or unsigned.不确定它是已签名还是未签名。 I was wondering how I can go about converting the hexadecimal to 38-bit integer to two 18-bit integers in python.
我想知道如何在 python 中将十六进制到 38 位整数转换为两个 18 位整数。
I saw in this post about converting hex to 16-bit signed integer using我在这篇文章中看到关于使用以下方法将十六进制转换为 16 位有符号整数
def twos_complement(hexstr,bits):
value = int(hexstr,16)
if value & (1 << (bits-1)):
value -= 1 << bits
return value
twos_complement('FFFE',16)
-2
Would this generalize to a 38-bit integer as well twos_complement('b3b4d25d0',38)
?这会推广到 38 位整数以及
twos_complement('b3b4d25d0',38)
吗?
For splitting an integer, I saw this post splitting a 16-bit int ( x
variable) to two 8-bit using为了拆分整数,我看到这篇文章将一个 16 位 int(
x
变量)拆分为两个 8 位使用
c = (x >> 8) & 0xff
f = x & 0xff
(1030333333 >> 8) & 0xff
163
(1030333333 >> 8)
4024739
Can this also be generalized to splitting a known 38-bit integer to two 16-bit integer using the method below?这也可以推广到使用下面的方法将一个已知的 38 位整数拆分为两个 16 位整数吗?
c = (x >> 16) & 0xff
f = x & 0xff
Assuming that a hex-string represents a positive integer you can just modify your last two lines into假设十六进制字符串表示一个正整数,您只需将最后两行修改为
>>> value = 0x50f116bcf
>>> msb = (value >> 18) & (2**18 - 1)
>>> lsb = value & (2**18 - 1)
# and convert (msb, lsb) back into value by using the binary OR
>>> (msb << 18) | lsb == value
True
or generalize this idea and use the following function或者概括这个想法并使用以下功能
def bin_grouper(value, bits):
"""bin_grouper(0b1011010, 3) --> 001 011 010"""
import math
num = math.ceil(value.bit_length() / bits) # number of blocks
mask = 2**bits - 1
blocks = [(value >> idx*bits) & mask for idx in range(num)]
fmtstr = f'{{:0{bits}b}}'
return [fmtstr.format(v) for v in reversed(blocks)]
in order split a 36-bit number into two 18-bit numbers:按顺序将一个 36 位数字拆分为两个 18 位数字:
>>> bin_grouper(0x50f116bcf, 18)
['010100001111000100', '010110101111001111']
# or
>>> bin_grouper(int('50f116bcf', 16), 18)
['010100001111000100', '010110101111001111']
In order to compare the result you can convert the 0x50f116bcf
into a 36-bit binary string and test the upper 18-bits (msb) and the lower 18-bits (lsb):为了比较结果,您可以将
0x50f116bcf
转换为 36 位二进制字符串并测试高 18 位 (msb) 和低 18 位 (lsb):
>>> bstr = f'{0x50f116bcf:036b}'
>>> bstr
'010100001111000100010110101111001111'
>>> blocks = bin_grouper(0x50f116bcf, 18)
>>> blocks[0] == bstr[:18] == f'{msb:018b}'
True
>>> blocks[1] == bstr[18:] == f'{lsb:018b}'
True
Here's a base converter:这是一个基本转换器:
def getbasetasifintisbaset(orig,basefrom=10, baseto=16):
num=orig
result = 0
i=0
while num != 0:
result += (num % basefrom) * (baseto ** i)
num //=basefrom
i += 1
return result
getbasetasifintisbaset(100,10,16)
#256
getbasetasifintisbaset(256,16,10)
#100
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.