比较正确的做法（应该）

用户4403728317

#1 Created at ... [Delete] [Delete and Lock User]

def createCounter():
	list1 = [0]
	def counter():
		list1[0] = list1[0] + 1
		return list1[0]
	return counter

孤月情风

#2 Created at ... [Delete] [Delete and Lock User]

为啥列表可以，直接赋值不行呢？

def createCounter():
	n = 0
	def counter():
		n = n + 1
		return n
	return counter

>>> nboundLocalError: local variable 'n' referenced before assignment

简简单单rsi

#3 Created at ... [Delete] [Delete and Lock User]

同问，求大神解答

水滴竹沙

#4 Created at ... [Delete] [Delete and Lock User]

个人理解新计算的n是局部变量，不能外层函数引用

def createCounter():

s=0

defcounter():

nonlocal s

s+=1

return s

return counter

涂山佳佳环理华Yang99599

#5 Created at ... [Delete] [Delete and Lock User]



# -*- coding: utf-8 -*-




def createCounter():

    r=0

    def counter():

        nonlocal r  #重点  意思和楼上差不多，再白话一点就是楼下参数的不知道r=0，nonlocal的作用是标识该变量是上一级函数中的局部变量

                          #不知这样理解是不是是否全对（理解有错还请指出）

        r+=1

        return r

        

    return counter




# 测试:

counterA = createCounter()

print(counterA(), counterA(), counterA(), counterA(), counterA()) # 1 2 3 4 5

counterB = createCounter()

if [counterB(), counterB(), counterB(), counterB()] == [1, 2, 3, 4]:

    print('测试通过!')