Convert hex string to int in Python

Question

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

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

Answer 1

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

x = int("deadbeef", 16)

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

>>> 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.)

Answer 2

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

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

Answer 3

For any given string s:

int(s, 16)

Answer 4

Convert hex string to int in Python

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

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

Both strings will suffice for conversion in this way:

>>> 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.

>>> int(string_1, 0)
65535

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

>>> 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:

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

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

>>> 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.

Answer 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.

print int(0xdeadbeef) # valid

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

Answer 6

The worst way:

>>> 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?

Answer 7

Or ast.literal_eval (this is safe, unlike eval):

ast.literal_eval("0xffff")

Demo:

>>> import ast
>>> ast.literal_eval("0xffff")
65535
>>> 

Answer 8

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

The following works for me:

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

Answer 9

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

Example

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

Answer 10

If you are using the python interpreter, you can just type 0x(your hex value) and the interpreter will convert it automatically for you.

>>> 0xffff

65535

Answer 11

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

For example

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