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Home/ Questions/Q 8360171
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T11:22:40+00:00 2026-06-09T11:22:40+00:00

Consider the following code snippet (Note I was using global because nonlocal keywords is

  • 0

Consider the following code snippet (Note I was using global because nonlocal keywords is not available in Python 2.7)

def foo(L,K):
    global count
    count = 0
    def bar(f,L):
        global count
        for  e in L:
            if e - f == K or f - e == K: count += 1
            yield e
    try:
        while True:
            L = bar(L.next(),L)
    except StopIteration:
        return count
count=0
print foo((int(e) for e in some_string.split()),some_number)

where

some_string: A space delimited integers
some_number: An integer

when len(some_string) = 4000, the above code fails with the error

RuntimeError: maximum recursion depth exceeded while calling a Python object

Is it because internally nested generators are implemented as recursion?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T11:22:43+00:00Added an answer on June 9, 2026 at 11:22 am

    You are replacing L, with the result of bar, which is a generator itself. Thus, you end up passing bar back to bar in the form of recursively nested generator expressions.

    This construction eventually passes the recursion depth limits.

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