Convert hex string to integer in Python

Question

How do I convert a hex string to an integer?

"0xffff"   ⟶   65535
"ffff"     ⟶   65535

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; if you omit the second parameter, int() will 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

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 4

For any given string s:

int(s, 16)

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

Please don't do this!

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

>>> hex_to_int("c0ffee")
12648430

Why is using 'eval' a bad practice?

15000+ examples of this in the wild.

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

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 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

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)