Discuss / Python / 3 solutions to createCounter()
Back

3 solutions to createCounter()

Topic source

KonoNA7

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

Solution 1:

def createCounter1():
    li = [0]
    def counter():
        li.append(li[-1] + 1)
        return li[-1]
    return counter

Solution 2:

def createCounter2():
    n = 0
    def counter():
        nonlocal n # 使用外层变量
        n += 1
        return n
    return counter

Solution 3:

def createCounter3():
    def f(): 
        n = 1
        while 1:
            yield n
            n += 1
    fn = f()
    def counter():
        return next(fn)
    return counter

KonoNA7

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

优化了一下Solution 1:

def createCounter1():
    li = [0]
    def counter():
        li[0] += 1
        return li[0]
    return counter

或者

def createCounter1():
    li = [1]
    def counter():
        li.insert(0, li[0] + 1)
        return li.pop()
    return counter

Carlos_EN

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

请问第一种方法里，调用后面的counterA()的时候，li为什么不会再定义为0？

KonoNA7

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

你可以理解为li = [0]只会在counterA = createCounter()赋值语句时执行，调用counterA()只执行counter()函数，并且可以使用li这个变量。

用户7041329703

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

我想问下，前面的 f1, f2, f3 为什么能带入；或者说这里的1、2、3怎么会带入到函数内的；fs又不是函数，它是list不是，list要带入不得这样的吗 f[1]；

谢谢，前面那个讲解也没看明白

KonoNA7

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

@用户7041329703

不好意思，我没能理解你的问题，可以阐述得再清楚一些吗:(

胖紫爱吃肉

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

不好意思，同样很疑惑在第一种方法里，为什么后续调用counterA()的时候Li不会重新赋值为0

KonoNA7

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

@胖紫爱吃肉

因为counterA指向的不是createCounter()，而是counter()：

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

li = [0]不在这个函数里面，后续再怎么调用counterA()都不会再执行这条语句。

胖紫爱吃肉

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

谢谢，我再自行理解下

小甜甜郑林200

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

同样的疑问，还是无法理解 在第一种方法里 在定义countA时， counterA = createCounter() 不就是理解为countA指向了counterA = createCounter()？ 而在createCounter()函数里应该是顺序执行 既然是顺序执行的话 在第二次第三次调用的时候难道没有重新定义li=[0]？ 求大神解释一下我在哪个环节理解错了

Reply