How to get hex string from signed integer

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?

From stackoverflow Ellery Newcomer

• 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:]
  

  Ellery Newcomer : .. or it would if I were less dyslexic ..

  bobince : Don't rely on the 'L' suffix, it is going away in Python 3.0 so hex2 will chop off a digit. The %x formatting operator is generally a better bet than hex().

  paxdiablo : Incorporated comment and Ignacios option below into accepted answer (and gave Ignacio an upvote)
  From paxdiablo

• '%#4x' % (-1 & 0xffffffff)
  

  From Ignacio Vazquez-Abrams