Convert hex string to integer in 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.)

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

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

For any given string s:

int(s, 16)

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

ast.literal_eval("0xffff")

Demo:

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

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

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)