Python Trailing L Problem

Question

I'm using Python to script some operations on specific locations in memory (32 bit addresses) in an embedded system.

When I'm converting these addresses to and from strings, integers and hex values a trailing L seems to appear. This can be a real pain, for example the following seemingly harmless code won't work:

int(hex(4220963601))

Or this:

int('0xfb96cb11L',16)

Does anyone know how to avoid this?

So far I've come up with this method to strip the trailing L off of a string, but it doesn't seem very elegant:

if longNum[-1] == "L":
   longNum = longNum[:-1]

Your first example wouldn't work even without the "L" because `int` doesn't know how to handle "0x" prefix. What are you trying to accomplish? — recursive, May 06 '11 at 21:33

score 24 · Accepted Answer · answered May 06 '11 at 21:32

24

If you do the conversion to hex using

 "%x" % 4220963601

there will be neither the 0x nor the trailing L.

answered May 06 '11 at 21:32

Sven Marnach

574,206
118
941
841

score 13 · Answer 2 · answered May 06 '11 at 21:32

13

Calling str() on those values should omit the trailing 'L'.

answered May 06 '11 at 21:32

yan

20,644
3
38
48

14

Just to let future readers know, yan's answer of calling 'str' on it doesn't work, at least in Python 2.7.3. Str keeps the trailing 'L'. (I don't have enough reputation to comment on Yan's answer) The answers using format or %x do seem to work. – Rhadamanthos Jan 07 '14 at 17:33
2

broken spacing, but: Python 2.7.5+ (default, Sep 19 2013, 13:48:49) [GCC 4.8.1] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> n = 123L >>> str(n) '123' >>> str(123L) '123' – yan Jan 07 '14 at 19:42
That example is true in general but for hex, hex does not accept strings it returns a string with L in it given an integer :'( – Har Apr 12 '17 at 16:25

score 3 · Answer 3 · answered Mar 09 '19 at 07:52

3

Consider using rstrip. For example:

result.rstrip("L")

answered Mar 09 '19 at 07:52

Conor Livingston

905
1
8
17

docpkd · Answer 4 · 2015-07-23T06:03:12.347

0

This is what I did: int(variable_which_is_printing_as_123L) and it worked for me. This will work for normal integers.

edited Jul 23 '15 at 06:03

answered Jul 23 '15 at 04:15

docpkd

551
9
19

score -1 · Answer 5 · edited Apr 21 '15 at 15:21

-1

this could help somebody:

>>>n=0xaefa5ba7b32881bf7d18d18e17d3961097548d7cL
>>>print "n=","%0s"%format(n,'x').upper()
n= AEFA5BA7B32881BF7D18D18E17D3961097548D7C

edited Apr 21 '15 at 15:21

kratenko

7,354
4
36
61

answered Apr 21 '15 at 15:12

tedy58-user1045284

19
2

3

how would this uncommented piece of code help anyone more than the answers given already? – kratenko Apr 21 '15 at 15:22

Python Trailing L Problem

5 Answers5

Linked

Related