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Home/ Questions/Q 655471
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T22:36:26+00:00 2026-05-13T22:36:26+00:00

I’m using Python 2.5. I’m trying to use this ‘with’ statement. from __future__ import

  • 0

I’m using Python 2.5. I’m trying to use this ‘with’ statement.

from __future__ import with_statement
a = []
with open('exampletxt.txt','r') as f:
    while True:
        a.append(f.next().strip().split())
print a

The contents of ‘exampletxt.txt’ are simple:

a
b

In this case, I get the error:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/tmp/python-7036sVf.py", line 5, in <module>
    a.append(f.next().strip().split())
StopIteration

And if I replace f.next() with f.read(), it seems to be caught in an infinite loop.

I wonder if I have to write a decorator class that accepts the iterator object as an argument, and define an __exit__ method for it?

I know it’s more pythonic to use a for-loop for iterators, but I wanted to implement a while loop within a generator that’s called by a for-loop… something like

def g(f):
    while True:
        x = f.next()
        if test1(x):
            a = x
        elif test2(x):
            b = f.next()
            yield [a,x,b]

a = []
with open(filename) as f:
    for x in g(f):
        a.append(x)
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  1. Editorial Team
    Editorial Team
    2026-05-13T22:36:27+00:00Added an answer on May 13, 2026 at 10:36 pm

    Raising StopIteration is what an iterator does when it gets to the end. Normally the for statement catches it silently and continues to the else clause, but if it’s being iterated manually as in your case then the code has to be prepared to handle the exception itself.

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