在Python中将十六进制字符串转换为int

How do I convert a hex string to an int in Python?

如何在Python中将十六进制字符串转换为int ?

I may have it as "0xffff" or just "ffff".

我可以把它写成“0xffff”或者只是“ffff”。

10 个解决方案

#1

Without the 0x prefix, you need to specify the base explicitly, otherwise there's no way to tell:

如果没有0x前缀，就需要显式地指定基数，否则无法判断:

x = int("deadbeef", 16)

With the 0x prefix, Python can distinguish hex and decimal automatically.

使用0x前缀，Python可以自动区分十六进制和十进制。

>>> print int("0xdeadbeef", 0)
3735928559
>>> print int("10", 0)
10

(You must specify 0 as the base in order to invoke this prefix-guessing behavior; omitting the second parameter means to assume base-10.)

(必须指定0为基数，才能调用这个前缀猜测行为;省略第二个参数表示假设为base-10。

#2

int(hexString, 16) does the trick, and works with and without the 0x prefix:

int(hexString, 16)就是这样做的，使用和不使用0x前缀:

>>> int("a", 16)
10
>>> int("0xa",16)
10

#3

For any given string s:

对于给定的字符串s:

int(s, 16)

#4

Convert hex string to int in Python

I may have it as "0xffff" or just "ffff".

我可以把它写成“0xffff”或者只是“ffff”。

To convert a string to an int, pass the string to int along with the base you are converting from.

要将字符串转换为int类型，将字符串传递给int类型以及要转换的基数。

Both strings will suffice for conversion in this way:

两个字符串都可以这样转换:

>>> string_1 = "0xffff"
>>> string_2 = "ffff"
>>> int(string_1, 16)
65535
>>> int(string_2, 16)
65535

Letting int infer

If you pass 0 as the base, int will infer the base from the prefix in the string.

如果您将0作为基数传递，int将从字符串中的前缀推断基数。

>>> int(string_1, 0)
65535

Without the hexadecimal prefix, 0x, int does not have enough information with which to guess:

没有十六进制前缀0x, int就没有足够的信息来猜测:

>>> int(string_2, 0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 0: 'ffff'

literals:

If you're typing into source code or an interpreter, Python will make the conversion for you:

如果您正在输入源代码或解释器，Python将为您进行转换:

>>> integer = 0xffff
>>> integer
65535

This won't work with ffff because Python will think you're trying to write a legitimate Python name instead:

这对ffff不起作用，因为Python会认为您正在尝试编写一个合法的Python名称:

>>> integer = ffff
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'ffff' is not defined

Python numbers start with a numeric character, while Python names cannot start with a numeric character.

Python数字以数字字符开头，而Python名称不能以数字字符开头。

#5

Adding to Dan's answer above: if you supply the int() function with a hex string, you will have to specify the base as 16 or it will not think you gave it a valid value. Specifying base 16 is unnecessary for hex numbers not contained in strings.

在上面添加Dan的答案:如果您使用十六进制字符串提供int()函数，那么您将不得不指定基数为16，否则它将不会认为您给了它一个有效值。对于字符串中不包含的十六进制数字，不需要指定基数为16。

print int(0xdeadbeef) # valid

myHex = "0xdeadbeef"
print int(myHex) # invalid, raises ValueError
print int(myHex , 16) # valid

#6

The worst way:

,最糟糕的莫过于:

>>> def hex_to_int(x):
    return eval("0x" + x)

>>> hex_to_int("c0ffee")
12648430

Please don't do this!

Is using eval in Python a bad practice?

在Python中使用eval是不好的实践吗?

#7

In Python 2.7, int('deadbeef',10) doesn't seem to work.

在Python 2.7中，int(“deadbeef”，10)似乎不起作用。

The following works for me:

以下是我的作品:

>>a = int('deadbeef',16)
>>float(a)
3735928559.0

#8

The formatter option '%x' % seems to work in assignment statements as well for me. (Assuming Python 3.0 and later)

格式化程序选项'%x' %似乎也适用于赋值语句。(假设Python 3.0及以上版本)

Example

例子

a = int('0x100', 16)
print(a)   #256
print('%x' % a) #100
b = a
print(b) #256
c = '%x' % a
print(c) #100

#9

with '0x' prefix, you might also use eval function

有了'0x'前缀，还可以使用eval函数

For example

例如

>>a='0xff'
>>eval(a)
255

#10

To convert a DWORD from hex to a signed integer , implement two's complement like this:

要将一个DWORD从十六进制转换为一个有符号整数，请实现如下的两个补码:

~ (0xffffffff - int('0xdeadbeef', 16)) + 1