Converting Hex To Int, The 'l' Character

I have a 64bit hex number and I want to convert it to unsigned integer. I run >>> a = 'ffffffff723b8640' >>> int(a,16) 18446744071331087936L So what is the 'L'

Solution 1:

Python2.x has 2 classes of integer (neither of them are unsigned btw). There is the usual class int which is based on your system's concept of an integer (often a 4-byte integer). There's also the arbitrary "precision" type of integer long. They behave the same in almost all circumstances and int objects automatically convert to long if they overflow. Don't worry about the L in the representation -- It just means your integer is too big for int (there was an Overflow) so python automatically created a long instead.

It is also worth pointing out that in python3.x, they removed python2.x's int in favor of always using long. Since they're now always using long, they renamed it to int as that name is much more common in code. PEP-237 gives more rational behind this decision.

Solution 2:

You are trying to apply string methods to an integer. But the string representation of a long integer doesn't have the L at the end:

In [1]: a = "ffffffff723b8640"

In [2]: int(a, 16)
Out[2]: 18446744071331087936L

In [3]: str(int(a, 16))
Out[3]: '18446744071331087936'

The __repr__ does, though (as @mgilson notes):

In [4]: repr(int(a, 16))
Out[4]: '18446744071331087936L'

In [5]: repr(int(a, 16))[:-1]
Out[5]: '18446744071331087936'