tahsmith tahsmith -4 years ago 82
Python Question

UnboundLocalError when using += on list. Why is `nonlocal` needed here when directly calling __iadd__ works fine?

Consider this code:

def main():
l = []

def func():
l += [1]

func()
print(l)

if __name__ == '__main__':
main()


It will produce:

Traceback (most recent call last):
File "/Users/tahsmith/Library/Preferences/PyCharm2017.1/scratches/scratch_21.py", line 14, in <module>
main()
File "/Users/tahsmith/Library/Preferences/PyCharm2017.1/scratches/scratch_21.py", line 11, in main
func()
File "/Users/tahsmith/Library/Preferences/PyCharm2017.1/scratches/scratch_21.py", line 9, in func
l += [1]
UnboundLocalError: local variable 'l' referenced before assignment


This itself can be fixed by either using
nonlocal l
at the start of
func
or using
__iadd__
directly instead of
+=
.

Question: Why is
nonlocal
needed here?

This is very surprising to me.

Answer Source

+= is the augmented assignment operator; it roughly translates to:

def func():
    l = l + [1]

If you were to replace l += [1] with a call to object.__iadd__(), you can't ignore the return value of that call if you were to use it properly:

def func():
    l = l.__iadd__([1])

Both of those translations also need a nonlocal statement, because both access l and assign back to l.

You can get away with ignoring the return value of object.__iadd__ because list objects are mutable; the list is mutated in-place. But then you may as well use the list.extend() call in that case:

def func():
    l.extend([1])

list.__iadd__(), under the covers, calls list.extend() before returning self.

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