递归函数没有返回? [英] Recursive function returning none?

问题描述

为了实现自己的二分查找，我编写了以下函数

I wrote the following function in order to implement my own binary search

def bisect(input, target):
    mid = len(input)/ 2
    if len(input) == 1:
        if input[0] == target:
            return 1
        else:
            return None
    elif input[mid] > target:
        bisect(input[:mid], target)
    elif input[mid] <= target:
        bisect(input[mid:], target)

我知道我的实现已经关闭，但我更想了解这里的递归堆栈.

I know my implementation is off, but I am more curious in understanding the recursive stack here.

当我调用 bisect(['d','e'], 'd') 时，我的函数应该返回

When I call bisect(['d','e'], 'd'), my function should return the value of

bisect(['d'], 'd')

而是返回 None.此外，当我直接调用 bisect(['d'], 'd') 时，我得到了正确的 0 值.这怎么可能?

but instead it returns None. Further, when I call bisect(['d'], 'd')directly, I get the correct value of 0. How is this possible?

推荐答案

您忽略了递归调用的返回值.你也需要明确返回那些:

You are ignoring the return values of recursive calls. You need to explicitly return those too:

elif input[mid] > target:
    return bisect(input[:mid], target)
elif input[mid] <= target:
    return bisect(input[mid:], target)

递归调用就像任何其他函数调用一样；他们将结果返回给调用者.如果你忽略返回值，然后调用函数结束，你最终会得到那个调用函数，然后返回 None.

Recursive calls are just like any other function call; they return a result to the caller. If you ignore the return value and the calling function then ends, you end up with that calling function then returning None instead.

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