Do I have to do StringIO.close()?

Question

Some code:

import cStringIO  def f():     buffer = cStringIO.StringIO()     buffer.write('something')     return buffer.getvalue() 

The documentation says:

StringIO.close(): Free the memory buffer. Attempting to do further operations with a closed StringIO object will raise a ValueError.

Do I have to do buffer.close(), or it will happen automatically when buffer goes out of scope and is garbage collected?

UPDATE:

I did a test:

import StringIO, weakref  def handler(ref):     print 'Buffer died!'  def f():     buffer = StringIO.StringIO()     ref = weakref.ref(buffer, handler)     buffer.write('something')     return buffer.getvalue()  print 'before f()' f() print 'after f()' 

Result:

vic@wic:~/projects$ python test.py  before f() Buffer died! after f() vic@wic:~/projects$ 

Answer 1

Generally it's still better to call close() or use the with statement, because there may be some unexpected behaviour in special circumstances. For example, the expat-IncrementalParser seems to expect a file to be closed, or it won't return the last tidbit of parsed xml until a timeout occurs in some rare circumstances.

But for the with-statement, which handles the closing for you, you have to use the StringIO class from the io-Modules, as stated in the comment of Ivc.

This was a major headache in some legacy sax-parser script we solved by closing the StringIO manually.

The "out-of-scope" close didn't work. It just waited for the timeout-limit.

Answer 2

From the source:

class StringIO:     ...     def close(self):         """Free the memory buffer.         """         if not self.closed:             self.closed = True             del self.buf, self.pos 

So StringIO.close just frees the memory buffer deleting references to StringIO.buf and StringIO.pos. But if self is garbage collected, its attributes will also be garbage collected, having the same effect as StringIO.close.