Calling context managers in python

Question

I am trying to find a good way to explain why creating an specialized context manager by calling another one works as expected. I am not complaining!, it is great that it does. But I was not sure if that it would work until I tested it and therefore I feel I am loosing something.

Just to clarify considere the following example:

>>> from contextlib import contextmanager
>>> @contextmanager
... def f(val):
...    print(val)
...    yield
...    print(val+1)
...
>>>
>>> with f(1):
...  print(3)
...
1
3
2

And now we specialize it:

>>> def f42():
...    return f(42)
...
>>> with f42():
...    print(3)
...
42
3
43

I guess what is confusing me is why does the yield from f bubbles up through f42? Why I do not need to write f42 as a context manager explicitly.

Martijn Pieters · Accepted Answer · 2014-10-13 14:38:25Z

1

There is no difference between using f(42) in a with statement and using it in another function then returning it. All that with requires is that the expression produces a context manager.

You could also do this:

cm = f(42)
with cm:
    print(3)

All that Python does is execute the expression in the with <expression> statement and then treats the result of that expression as a context manager. How the expression produced that context manager is of no consequence.

In other words, f is not your context manager here, only the return value of a call to f().

answered Oct 13, 2014 at 14:38

Martijn Pieters

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ErlVolton · Accepted Answer · 2014-10-13 14:36:25Z

1

In your last with statement, f42() gets evaluated and runs f(42) which is then used in the context of your with statement.

answered Oct 13, 2014 at 14:36

ErlVolton

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