Python 循环内的递归调用。它是如何工作的？

Question

我遇到了一个内部有递归函数调用的循环，其中循环的起始范围按如下方式递增。该代码输出以下序列。但是，我无法概念化为什么会生成这个特定序列。有人可以对其工作原理有一些见解吗？以及将这个递归函数转换为输出相同序列的迭代函数有多可行。请帮忙。

代码:

def foo(step=0):
    for i in range(step, 4):
        print step
        foo(step+1)

foo()   

输出:

0 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3 0 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3 0 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3 0 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3

类似设计的代码来查找 Anagrams:

def find_anagrams(word, step=0):
    print 'step->', step
    if step == len(word):
        print "".join(word)
    for i in range(step, len(word)):
        print step, i
        word_ = list(word)
        word_[step], word_[i] = word_[i], word_[step]
        find_anagrams(word_, step+1)

Answer 1

让我试试:

从您的代码片段中，在每个函数调用中，即 foo(step+1) ，都有一个称为激活记录的结构或 创建框架是为了存储有关函数调用进度的信息。 因此，当函数的执行导致嵌套函数调用时，前一个调用的执行 被挂起，它的激活记录存储了源代码中控制流的位置 嵌套调用返回后应继续。

主要部分如下:

当step == 4，即range(4,4) ==空列表时，该时间迭代不会发生，因此它将返回 没有任何。然后它将移动到之前停止的帧并开始新的迭代和递归函数调用 直到范围(4,4)。

注意:递归基本情况仅当步骤 == 4 时，即时间范围(4,4)并返回 None 。

每个递归都需要一个基本情况，否则它将进入无限循环。

那么，让我们看看递归跟踪:我添加了 i 来区分 step 和迭代增量。

# 1 def foo(step=0):
# 2    for i in range(step, 4):
# 3        print 'i: %d, step: %d' % (i,step)
# 4        foo(step+1)
# 5 foo()

line 5
line 1  foo with step=0  which is default
line 2  range(0,4)                       Frame: A,   0 is over, next=1
line 3  step = 0            Output: i: 0, step: 0
line 4  calling foo(0 + 1)
line 1  foo with step=1
line 2  range(1,4)                       Frame: B,   1 is over, next=2
line 3  step = 1            Output: i: 1, step: 1
line 4  calling foo(1 + 1)
line 1  foo with step=2
line 2  range(2,4)                       Frame: C,   2 is over, next=3
line 3  step = 2            Output: i: 2, step: 2
line 4  calling foo(2 + 1)
line 1  foo with step=3
line 2  range(3,4)                       Frame: D,   3 is over, next=4
line 3  step = 3,           Output: i: 3, step: 3
line 4  calling foo(3 + 1)
line 1  foo with step=4
line 2  range(4,4)                       Frame: E,
         This is an empty list, so it won't come inside the loop, so return None.
         Come back to previous Frame: D, i=3 was used, now increment to 4. So, again range(4,4)
line 2  range(4,4)          Empty list, from Frame: D, return None
         Come back to previous Frame C, now i=3, step was called with value 2
line 2  range(2,4)
line 3  step = 2            Output: i: 3, step: 2, why step == 2 because the function foo was called with step=2
line 4  calling foo(2 + 1)
line 1  foo with step=3
line 2  range(3,4)
line 3  step = 3,           Output : i: 3, step: 3
line 4  calling foo(3 + 1)
line 1  foo with step=4
line 2  range(4,4)          Empty list again, not going inside the list, return None
line 2  range(2,4)          From Frame: B, step was == 1, because the function foo was called with step=1
line 3  step: 1             Output: i: 2, step: 1,  here i ==2, because this is the second iteration of Frame B.
line 4  calling foo(1 + 1)
line 1  foo with step=2
line 2  range(2,4)
line 3  step: 2            Output: i: 2, step: 2

此后，它遵循相同的递归方式，直到迭代范围耗尽，即 range(4,4)

如果有帮助，请告诉我。