Taking a generator out of a context manager

Question

I just saw the following code:

from __future__ import print_function
from future_builtins import map # generator

with open('test.txt', 'r') as f:
    linegen = map(str.strip, f)

# file handle should be closed here

for line in linegen:
    # using the generator now
    print(line)

What happens in this case? Is the context manager smart enough to know that linegen still has a reference to the file handle, such that it's not closed when the context is left? Or is this potentially unsafe?

Answer 1

This is one of those breaking changes in Python 3.

Your question title (Taking a generator ...) implies you are reading it as Python 3 code.

But the statement

from __future__ import print_function

implies it was written for Python 2

In Python 2, map returns an actual list - and thus this code is both perfectly safe and extrmely sensible (i.e open a file, read all lines,s tripping them as you go, then close the file)

In [2]: with open('README.md','r') as f:
   ...:     lines = map(str.strip, f)
   ...:     
In [3]: lines
Out[3]: 
['ASM',
 '============',
 '',

In Python 3, the same code throws an exception

In [1]: with open('README.md','r') as f:
    lines = map(str.strip, f)
   ...:     
In [2]: lines
Out[2]: <map at 0x7f4d393c3ac8>
In [3]: list(lines)
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-3-2666a44c63b5> in <module>()
----> 1 list(lines)

ValueError: I/O operation on closed file.

If you want a version-safe implementation of this you need to either convert the generator to a list

lines = list(map(str.strip, f))

or just use a list comprehension

lines = [l.strip() for l in f]