Question

我有这个简单的功能：

def double(x)   : return x*2
def triple(x)   : return x*3
def increment(x): return x+1

def find(n, moves=[]):
    if n == 1:
        print("Inside if statement with return")
        print("`moves` is not None: moves=", moves)
        return moves
    if n % 2 == 0:
        find(n // 2, moves + [double])
    elif n % 3 == 0:
        find(n // 3, moves + [triple])
    else:
        find(n - 1 , moves + [increment])


print("find has returned: ", find(23))

我希望函数列表作为返回值，但实际输出是：

Inside if statement with return
`moves` is not None: moves= [<function increment at 0x7f03e3944598>, <function double at 0x7f03e9f6cbf8>, <function increment at 0x7f03e3944598>, <function double at 0x7f03e9f6cbf8>, <function increment at 0x7f03e3944598>, <function double at 0x7f03e9f6cbf8>, <function double at 0x7f03e9f6cbf8>]
find has returned:  None

这是非常意外的，因为moves在最终None语句中不是if，如print语句所示，但是返回None。

在[:]未更改任何内容后添加return moves，仍会返回None。

我如何return moves的实际内容？

包含函数的列表返回为None

0 个答案: