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Comment by mdaniel | Hacker News Reader

I didn't know the answer to this, but it seems like this actually behaves closer to `funcs.append("""print(i)""")` since

    funcs = list()
    for i in range(10):
        funcs.append(lambda: print(i))
        del i
    for f in funcs:
        f()

actually NameErrors during f() so I don't think it is actually closing over the i as much as the i was left in scope and the code was seemingly lazy evaluated

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Python has real closures. In my example, it is closing over i. In your example, you're deleting the closed-over variable. It's a bit weird, and Python is the only language that I know of that lets you do this.

Right. In this example lambda does not capture i at all. You would need something like

    funcs.append(lambda i=i: print(i))

No, it definitely captures i

Maybe this example where it returns the closures from a function is convincing?

    def fs():
        funcs = list()
        for i in range(10):
            funcs.append(lambda: print(i))
        return funcs
   
    for f in fs():
        f()

prints all 9s.

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