Convert hex string to int in Python
PythonStringHexPython Problem Overview
How do I convert a hex string to an int in Python?
I may have it as "0xffff" or just "ffff".
Python Solutions
Solution 1 - Python
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; if you omit the second parameter int() will assume base-10.)
Solution 2 - Python
int(hexstring, 16) does the trick, and works with and without the 0x prefix:
>>> int("a", 16)
10
>>> int("0xa", 16)
10
Solution 3 - Python
> # 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.
Solution 4 - Python
For any given string s:
int(s, 16)
Solution 5 - Python
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
Solution 6 - Python
The worst way:
>>> def hex_to_int(x):
return eval("0x" + x)
>>> hex_to_int("c0ffee")
12648430
#Please don't do this!#
https://stackoverflow.com/questions/1832940/is-using-eval-in-python-a-bad-practice
Solution 7 - Python
Or ast.literal_eval (this is safe, unlike eval):
ast.literal_eval("0xffff")
Demo:
>>> import ast
>>> ast.literal_eval("0xffff")
65535
>>>
Solution 8 - Python
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
Solution 9 - Python
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
Solution 10 - Python
Handles hex, octal, binary, int, and float
Using the standard prefixes (i.e. 0x, 0b, 0, and 0o) this function will convert any suitable string to a number. I answered this here: https://stackoverflow.com/a/58997070/2464381 but here is the needed function.
def to_number(n):
''' Convert any number representation to a number
This covers: float, decimal, hex, and octal numbers.
'''
try:
return int(str(n), 0)
except:
try:
# python 3 doesn't accept "010" as a valid octal. You must use the
# '0o' prefix
return int('0o' + n, 0)
except:
return float(n)
Solution 11 - Python
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
Solution 12 - Python
with '0x' prefix, you might also use eval function
For example
>>a='0xff'
>>eval(a)
255