How do I write a decorator that restores the cwd?

Question

How do I write a decorator that restores the current working directory to what it was before the decorated function was called? In other words, if I use the decorator on a function that does an os.chdir(), the cwd will not be changed after the function is called.

I had written the code and then it turned out (after refactoring) that I didn't need it. I figured stackoverflow is a good place to archive it, and perhaps others can benefit. — Daryl Spitzer, Oct 03 '08 at 23:31
I think the "Jeopardy" comment is just supposed to mean the title should still be "how do I write a decorator for X?", not "here's useful decorator for X". I agree that a note in the question would save people wasting time on duplicate solutions while you're typing. But they might better yours... — Steve Jessop, Oct 04 '08 at 14:33
... for instance codeape's point about exception-handling is an improvement on the questioner's proposed solution in this case, so it wasn't time wasted. — Steve Jessop, Oct 04 '08 at 14:37
Well I'm glad I asked the question, because codeape improved upon my answer, and ΤΖΩΤΖΙΟΥ came up with a context manager which didn't occur to me. — Daryl Spitzer, Oct 08 '08 at 00:49
Context manager is a better choice than decorator. Modern answer: stdlib [`contextlib.chdir`](https://docs.python.org/3/library/contextlib.html#contextlib.chdir). — wim, May 27 '23 at 01:47

tzot · Answer 1 · 2008-10-04T18:56:41.387

56

The answer for a decorator has been given; it works at the function definition stage as requested.

With Python 2.5+, you also have an option to do that at the function call stage using a context manager:

from __future__ import with_statement # needed for 2.5 ≤ Python < 2.6
import contextlib, os

@contextlib.contextmanager
def remember_cwd():
    curdir= os.getcwd()
    try: yield
    finally: os.chdir(curdir)

which can be used if needed at the function call time as:

print "getcwd before:", os.getcwd()
with remember_cwd():
    walk_around_the_filesystem()
print "getcwd after:", os.getcwd()

It's a nice option to have.

EDIT: I added error handling as suggested by codeape. Since my answer has been voted up, it's fair to offer a complete answer, all other issues aside.

edited Oct 04 '08 at 18:56

answered Oct 03 '08 at 22:19

tzot

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*And* it can be used to write the aforementioned decorator :) – Constantin Oct 03 '08 at 23:25
Does the error handling need an explicit try/finally? I thought the point of context managers was that MANAGER.__exit__ was always called. But then, I have never tried the decorator from contextlib, so I don't know what the issues are. – Adrian Ratnapala Oct 21 '11 at 06:22
No, it doesn't need an explicit `try/finally`, but you most probably want a `try/except` clause to handle failures. – tzot Oct 21 '11 at 07:47

score 37 · Accepted Answer · edited Dec 26 '17 at 15:41

37

The path.py module (which you really should use if dealing with paths in python scripts) has a context manager:

subdir = d / 'subdir' #subdir is a path object, in the path.py module
with subdir:
  # here current dir is subdir

#not anymore

(credits goes to this blog post from Roberto Alsina)

edited Dec 26 '17 at 15:41

Hyunjun Kim

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answered Dec 24 '12 at 09:38

CharlesB

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If path.py is now built-in, perhaps you should answer http://stackoverflow.com/questions/3899761/will-the-real-path-py-please-stand-up. – Daryl Spitzer Dec 27 '12 at 00:50
unfortunately it isn't, but thanks I wasn't aware of the question – CharlesB Dec 27 '12 at 05:24
I guess I misinterpreted your answer. Did you mean that the context manager is built in to path.py? (I thought you meant path.py is now built in to Python.) – Daryl Spitzer Dec 27 '12 at 18:05
4

Because Daryl Spitzer assumed that `path.py` is the same as `pathlib`, you may also assume that their context managers behave in same way, but they aren't: `pathlib.Path` doesn't touch CWD and instead acts as a context manager from `open()` ("closes" potentially open file descriptor). – Pugsley Oct 03 '20 at 06:25

score 26 · Answer 3 · edited May 08 '15 at 18:49

The given answers fail to take into account that the wrapped function may raise an exception. In that case, the directory will never be restored. The code below adds exception handling to the previous answers.

as a decorator:

def preserve_cwd(function):
    @functools.wraps(function)
    def decorator(*args, **kwargs):
        cwd = os.getcwd()
        try:
            return function(*args, **kwargs)
        finally:
            os.chdir(cwd)
    return decorator

and as a context manager:

@contextlib.contextmanager
def remember_cwd():
    curdir = os.getcwd()
    try:
        yield
    finally:
        os.chdir(curdir)

score 14 · Answer 4 · edited Mar 07 '23 at 18:39

14

You dont need to write it for you. With python 3.11, the developers have written it for you. Check out their code at github.com.

import contextlib
with contextlib.chdir('/path/to/cwd/to'):
    pass

edited Mar 07 '23 at 18:39

Martin Valgur

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answered May 08 '22 at 17:24

Machinexa

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score 3 · Answer 5 · answered Oct 03 '08 at 22:06

def preserve_cwd(function):
   def decorator(*args, **kwargs):
      cwd = os.getcwd()
      result = function(*args, **kwargs)
      os.chdir(cwd)
      return result
   return decorator

Here's how it's used:

@preserve_cwd
def test():
  print 'was:',os.getcwd()
  os.chdir('/')
  print 'now:',os.getcwd()

>>> print os.getcwd()
/Users/dspitzer
>>> test()
was: /Users/dspitzer
now: /
>>> print os.getcwd()
/Users/dspitzer

How do I write a decorator that restores the cwd?

5 Answers5

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