在python中从内部函数修改函数变量

Question

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)?

Answer 1

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.

Answer 2

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).

Answer 3

Workaround for Python 2:

def outer():
    a = [1]
    def inner():
        a[0] += 1
        print a[0]