闭包函数

闭包函数就是返回值是函数的函数

# _*_coding:utf-8 _*_

#求和函数lazy_sum（）的返回值是求和函数sum(),所以需要再次调用sum（）函数才能算出结果
def lazy_sum(*args):
    def sum():
        ax = 0
        for n in args:
            ax += n
        return ax

    return sum


L = lazy_sum(2, 3, 4, 5)()
print(L)

#返回函数不要引用任何循环变量，或者后续会发生的变量。需要用到循环时，应该把循环放在返回函数之外
def count():
    def f(j):
        def g():
            return j * j

        return g

    fs = []
    for i in range(1, 4):
        fs.append(f(i))
    return fs


f1, f2, f3 = count()
print(f1())
print(f2())
print(f3())


# 练习：利用闭包返回一个计数器函数，每次调用它返回递增整数：

# 解法1 定义一些列表，往里面不断的添加最后一个元素+1的值
def createCounter1():
    L = [0]

    def counter():
        i = L[-1]
        L.append(i + 1)
        return L[-1]

    return counter


counterA = createCounter1()

if [counterA(), counterA(), counterA(), counterA()] == [1, 2, 3, 4]:
    print('pass')
else:
    print('failed')


# 解法2 创建只有一个元素的列表，每次替换这个元素
def createCounter2():
    L = [0]

    def counter():
        L[0] += 1

    return L[0]


counterB = createCounter1()
if [counterB(), counterB(), counterB(), counterB()] == [1, 2, 3, 4]:
    print('pass')
else:
    print('failed')


# 解法3：创造一个生产器，每次调用函数时调用生成器来取值
def createCounter3():
    def f():
        n = 0
        while True:
            n += 1
            yield n

    sum = f()

    def counter():
        return (next(sum))

    return counter


counterC = createCounter3()
if [counterC(), counterC(), counterC(), counterC()] == [1, 2, 3, 4]:
    print('pass')
else:
    print('failed')