in reply to Re^9: Closures and scope of lexicals
in thread Closures and scope of lexicals

Your code was also over complicating Python, you never needed the variable i, because j never goes out of scope. 

subrefs = []

for j in range(1,4):
   def subref():
      print( j )

   subrefs.append( subref )

for subref in subrefs:
   subref()

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3
3
3

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Cheers Rolf
(addicted to the Perl Programming Language :)
see Wikisyntax for the Monastery

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Re^11: Closures and scope of lexicals
by ikegami (Patriarch) on Oct 31, 2024 at 17:51 UTC

    The OP's entire question is about the value of i. You can't just remove it! Chill out, man. You made a mistake, fine, but now you're compounding it time after time.

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