问题
I would like to test the example of the use of the nonlocal statement specified in the answer on this question:
def outer():
x = 1
def inner():
nonlocal x
x = 2
print("inner:", x)
inner()
print("outer:", x)
but when I try to load this code, I always get a syntax error:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "t.py", line 4
nonlocal x
^
SyntaxError: invalid syntax
Does anybody know what I am doing wrong here (I get the syntax error for every example that I use, containing nonlocal
).
回答1:
nonlocal
only works in Python 3; it is a new addition to the language.
In Python 2 it'll raise a syntax error; python sees nonlocal
as part of an expression instead of a statement.
This specific example works just fine when you actually use the correct Python version:
$ python3.3
Python 3.3.0 (default, Sep 29 2012, 08:16:08)
[GCC 4.2.1 Compatible Apple Clang 3.1 (tags/Apple/clang-318.0.58)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def outer():
... x = 1
... def inner():
... nonlocal x
... x = 2
... print("inner:", x)
... inner()
... print("outer:", x)
...
回答2:
Names listed in a nonlocal statement must not collide with pre-existing bindings in the local scope.
https://docs.python.org/3/reference/simple_stmts.html#the-nonlocal-statement
def outer():
x = 1
def inner():
nonlocal x
y = 2
x = y
print("inner: ", x)
inner()
print("outer: ", x)
>>> outer()
inner: 2
outer: 2
来源:https://stackoverflow.com/questions/14264313/syntax-error-on-nonlocal-statement-in-python