在Python中将十六进制值转换为十进制

Question

I just started working with hexadecimal values in python and am a bit surprised with what I just encountered. I expected the following code to first print a hexadecimal string, and then a decimal value.

Input:

n = 0x8F033CAE74F88BA10D2BEA35FFB0EDD3
print('Hex value for n is:', n)
print('Dec value for n is:', int(str(n), 16))

Output:

Hex value for n is: 190096411054295805012706659640261275091

Dec value for n is: 8921116140846515089057635273465667902228615313

How is it possible that 2 different different numbers are shown? I expected the first number to be a hexadecimal string and the second it's decimal equivalent, what is this second value in this case?

Answer 1

0x is a way to input an integer with an hexadecimal notation.

>>> n = 0x8F033CAE74F88BA10D2BEA35FFB0EDD3

This hexadecimal notation is forgotten directly after instantiation, though:

>>> n
190096411054295805012706659640261275091
>>> str(n)
'190096411054295805012706659640261275091'

So when you call int(str(n), 16) , Python interprets '190096411054295805012706659640261275091' as an hexadecimal number:

>>> int(str(n), 16)
8921116140846515089057635273465667902228615313

You need to input the original hex string:

>>> int("8F033CAE74F88BA10D2BEA35FFB0EDD3", 16)
190096411054295805012706659640261275091

or use hex :

>>> int(hex(n), 16)
190096411054295805012706659640261275091