It's ok to get and print the outer function variable a
def outer():
a = 1
def inner():
print a
It's also ok to get the outer function array a
and append something
def outer():
a = []
def inner():
a.append(1)
print a
However, it caused some trouble when I tried to increase the integer:
def outer():
a = 1
def inner():
a += 1 #or a = a + 1
print a
>> UnboundLocalError: local variable 'a' referenced before assignment
Why does this happen and how can I achieve my goal (increase the integer)?
In Python 3 you can do this with the nonlocal
keyword. Do nonlocal a
at the beginning of inner
to mark a
as nonlocal.
In Python 2 it is not possible.
A generally cleaner way to do this would be:
def outer():
a = 1
def inner(b):
b += 1
return b
a = inner(a)
Python allows a lot, but non-local variables can be generally considered as "dirty" (without going into details here).
Workaround for Python 2:
def outer():
a = [1]
def inner():
a[0] += 1
print a[0]
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.