Alexander Myodov wrote:
> So, with 2.5, I tried to utilize "with...as" construct for this, but
> unsuccessfully:
> from __future__ import with_statement
> with 5 as k:
> pass
> print k
> - told me that "AttributeError: 'int' object has no attribute
> '__context__'".
>
>
> So, does this mean that we still don't have any kind of RIIA in
> Python, any capability to localize the lifetime of variables on a
> level less than a function, and this is indeed not gonna happen to
> change yet?
>
No, it means that Python 2.5 supports 'resource initialisation is
acquisition', but that has nothing to do with the restricting the lifetime
of a variable. You have to use a context manager to handle the resource, 5
isn't a context manager. Some objects which actually need handling as a
resource can be used as context managers, for others you might need to
write your own.
with open('/etc/passwd', 'r') as f:
for line in f:
print line
after executing this f has been closed, but the variable f still exists.
Or try this:
>>> @contextlib.contextmanager
... def silly(n):
... print "starting to use", n
... yield n
... print "finished with", n
...
>>> with silly(5) as k:
... print "hello"
...
starting to use 5
hello
finished with 5
>>> k
5
The resource is controlled by the with statement, but the scope of the
variable and the lifetime of the object are separate issues.
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