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If __debug__ is present in globals(), then the globals() value is read instead of the actual __debug__ value. However, the write is still attempted to the real __debug__, which fails. (or maybe it failed early when it detected a write to __debug__)
If
__debug__
is present inglobals()
, then theglobals()
value is read instead of the actual__debug__
value. However, the write is still attempted to the real__debug__
, which fails. (or maybe it failed early when it detected a write to__debug__
)The text was updated successfully, but these errors were encountered: