How to get hex string from signed integer

Question

Say I have the classic 4-byte signed integer, and I want something like

print hex(-1)

to give me something like

0xffffffff

In reality, the above gives me -0x1 . I'm dawdling about in some lower level language, and python commandline is quick n easy.

So.. is there a way to do it?

Answer 1

This will do the trick:

>>> print hex (-1 & 0xffffffff)
0xffffffffL

or, in function form (and stripping off the trailing "L"):

>>> def hex2(n):
...     return hex (n & 0xffffffff)[:-1]
...
>>> print hex2(-1)
0xffffffff
>>> print hex2(17)
0x11

or, a variant that always returns fixed size (there may well be a better way to do this):

>>> def hex3(n):
...     return "0x%s"%("00000000%s"%(hex(n&0xffffffff)[2:-1]))[-8:]
...
>>> print hex3(-1)
0xffffffff
>>> print hex3(17)
0x00000011

Or, avoiding the hex() altogether, thanks to Ignacio and bobince:

def hex2(n):
    return "0x%x"%(n&0xffffffff)

def hex3(n):
    return "0x%s"%("00000000%x"%(n&0xffffffff))[-8:]

Answer 2

试试这个功能：

'%#4x' % (-1 & 0xffffffff)

Answer 3

"0x{:04x}".format((int(my_num) & 0xFFFF), '04x') ，其中 my_num 是所需的数字