Python调用引发异常，而不是在递归函数中返回 [英] Python Calling a raise exception instead of returning in recursive function

问题描述

我必须检查在抛出异常以结束递归而不是返回的代码中，代码能以多快的速度工作

I have to check how much faster works a code where I throw exception to end a recursion instead of return

我的返回代码如下:

    def factorial(self, n):
    if n==0:
        return 1
    else:
        return n* self.factorial(n-1)

现在我的代码加注:

def factorial2(self, n):
    if n==0:
        raise Exc(1)
    else:
        return n* self.factorial2(n-1)

我不知道，是否应该这样看，我应该怎么做才能获得价值 我已经尝试过

I don't know, if it should look this way and what should I do to get a value I've tried sth like this

try:
    print factorial2(n)
except Exc:
    1

但是它不起作用.我什至不知道我是否接近好的解决方案.

But it doesn't work. I don't even know if I'm close to good solution.

有人可以帮我，给我一些提示吗?

Can anybody help me, give me some tips?

推荐答案

您需要在此函数的else部分中使用try/except ...因为如果您在外面听，所有调用最终都会引发此异常，则您的所有工作都是丢失

you need the try/except in the else part of this function ... since all calls will eventually raise this exception if you listen outside then all your work is lost

def factorial2(self, n):
    if n==0:
        raise Exc(1)
    else:
        try:
           return n* self.factorial2(n-1)
        except Exc:
           return n

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