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简单的改一下就可以了

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J0sk1

#1 Created at ... [Delete] [Delete and Lock User] 
def count():
    fs = []
    for i in range(1, 5):
        def f():
             return i*i
        fs.append(f())#此处调用f()就可以了
    return fs
f1 ,f2 ,f3 ,f4 = count()
print(f1,f2,f3,f4)
#输出 1,4,9,16

NEWWAVE微博起的名真2B

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

你这个返回的已经不是函数列表了

笑劫戈和囧囧眉

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

这个结果正确 ，但是并不是闭包结果 print(f1)可以直接输出结果1 而不是print(f1())才能输出结果

看彩虹的人

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

结果是对的，但是这个改动已经失去意义了，f（）是个确定的值了，返回的fs=【1,4,9】.没意义了。

yedihh

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

f1 ,f2 ,f3 ,f4 = count()执行时，fs列表的结果已经计算出来了，分别赋值给f1 ,f2 ,f3 ,f4，是这样理解吧

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