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上面是源代码
def cycle(f1, f2, f3):
"""Returns a function that is itself a higher-order function.
>>> def add1(x):
... return x + 1
>>> def times2(x):
... return x * 2
>>> def add3(x):
... return x + 3
>>> my_cycle = cycle(add1, times2, add3)
>>> identity = my_cycle(0)
>>> identity(5)
5
>>> add_one_then_double = my_cycle(2)
>>> add_one_then_double(1)
4
>>> do_all_functions = my_cycle(3)
>>> do_all_functions(2)
9
>>> do_more_than_a_cycle = my_cycle(4)
>>> do_more_than_a_cycle(2)
10
>>> do_two_cycles = my_cycle(6)
>>> do_two_cycles(1)
19
""""*** YOUR CODE HERE ***"
def h1(n):
def h2(x):
while n > 3:
x = f3(f2(f1(x)))
n -= 3
if n == 0:
x =x
elif n == 1:
x = f1(x)
elif n == 2:
x = f2(f1(x))
elif n == 3:
x = f3(f2(f1(x)))
return x
return h2
return h1
这样写的话
>>> def add1(x):
... return x + 1
>>> def times2(x):
... return x * 2
>>> def add3(x):
... return x + 3
>>> my_cycle = cycle(add1, times2, add3)
>>> identity = my_cycle(0)
>>> identity(5)
而后会报一个
UnboundLocalError: local variable ‘n’ referenced before assignment
这样的谬误
然而如果改成这样
def cycle(f1, f2, f3):
"""Returns a function that is itself a higher-order function.
>>> def add1(x):
... return x + 1
>>> def times2(x):
... return x * 2
>>> def add3(x):
... return x + 3
>>> my_cycle = cycle(add1, times2, add3)
>>> identity = my_cycle(0)
>>> identity(5)
5
>>> add_one_then_double = my_cycle(2)
>>> add_one_then_double(1)
4
>>> do_all_functions = my_cycle(3)
>>> do_all_functions(2)
9
>>> do_more_than_a_cycle = my_cycle(4)
>>> do_more_than_a_cycle(2)
10
>>> do_two_cycles = my_cycle(6)
>>> do_two_cycles(1)
19
""""*** YOUR CODE HERE ***"
def h1(num):
def h2(x):
n = num
while n > 3:
x = f3(f2(f1(x)))
n -= 3
if n == 0:
x =x
elif n == 1:
x = f1(x)
elif n == 2:
x = f2(f1(x))
elif n == 3:
x = f3(f2(f1(x)))
return x
return h2
return h1
就没有问题,阐明在里面函数中定义的这个 num
是能够拿到的,然而却不能够间接用,最最最奇怪的是,看到答案这么写
def ret_fn(n):
def ret(x):
i = 0
while i < n:
if i % 3 == 0:
x = f1(x)
elif i % 3 == 1:
x = f2(x)
else:
x = f3(x)
i += 1
return x
return ret
return ret_fn
这样写没有问题,阐明这个 n 既能够拿到,也能够间接用,我人都傻了
正文完