即使它不调用自身，该函数是否也是递归的? [英] Is this function recursive even though it doesn't call itself?

问题描述

from pythonds.basic.stack import Stack

rStack = Stack()

def toStr(n,base):
    convertString = "0123456789ABCDEF"
    while n > 0:
        if n < base:
            rStack.push(convertString[n])
        else:
            rStack.push(convertString[n % base])
        n = n // base
    res = ""
    while not rStack.isEmpty():
        res = res + str(rStack.pop())
    return res

print(toStr(1345,2))

我指的是本教程并粘贴了代码以上.该教程说该函数是递归的，但我在任何地方都没有看到递归调用，只是一个 while 循环.我错过了什么?

I'm referring to this tutorial and also pasted the code above. The tutorial says the function is recursive but I don't see a recursive call anywhere, just a while loop. What am I missing?

推荐答案

这个特定函数不是递归是对的.然而，上下文是，在上一张幻灯片上有一个递归函数，在这张幻灯片中，他们想一瞥它的内部行为.他们后来说:

You are right that this particular function is not recursive. However, the context is, that on the previous slide there was a recursive function, and in this one they want to show a glimpse of how it behaves internally. They later say:

前面的例子 ^{[即有问题的那个 - B.]} 让我们深入了解 Python 如何实现递归函数调用.

The previous example ^{[i.e. the one in question - B.]} gives us some insight into how Python implements a recursive function call.

所以，是的，标题具有误导性，它应该是扩展递归函数或用堆栈模仿递归函数行为或类似的东西.

So, yes, the title is misleading, it should be rather Expanding a recursive function or Imitating recursive function behavior with a stack or something like this.

有人可能会说这个函数在某种意义上采用递归方法/策略来解决正在解决的问题，但它本身并不是递归的.

One may say that this function employs a recursive approach/strategy in some sense, to the problem being solved, but is not recursive itself.

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