Issue 24321: interaction of nonlocal and except leading to incorrect behavior (original) (raw)

Issue24321

Created on 2015-05-29 01:55 by whitequark, last changed 2022-04-11 14:58 by admin. This issue is now closed.

Messages (3)
msg244356 - (view)	Author: whitequark (whitequark)	Date: 2015-05-29 01:55
To reproduce in Python 3.4.2: def f(): x = None def g(): nonlocal x try: raise Exception() except Exception as x: pass g() # ↓ UnboundLocalError: local variable 'x' referenced before assignment print("x", x) f() Compare this to: def f(): x = None def g(): nonlocal x with open("/dev/null") as x: pass g() print("x", x) f() (which prints: x <_io.TextIOWrapper name='/dev/null' mode='r' encoding='UTF-8'>)
msg244357 - (view)	Author: Martin Panter (martin.panter) *	Date: 2015-05-29 04:01
The first example seems to behave as I would expect. The UnboundLocalError is raised by the print() call, because the “x” variable has been deleted by the exception handler. Equivalent code without using “nonlocal”: >>> def f(): ... x = None ... try: ... raise Exception() ... except Exception as x: ... pass ... print("x", x) # UnboundLocal due to exception handler ... >>> f() Traceback (most recent call last): File "", line 1, in File "", line 7, in f UnboundLocalError: local variable 'x' referenced before assignment In both cases, I think this is correct behaviour. See <https://docs.python.org/3.4/reference/compound_stmts.html#except>, which says “When an exception has been assigned using ‘as target’, it is cleared at the end of the except clause.”
msg244368 - (view)	Author: R. David Murray (r.david.murray) *	Date: 2015-05-29 11:55
Indeed, if you replace the except clause with a 'del x', you get the same UnboundLocalError. This is working as designed.