遞歸 function 在下面的示例中如何工作？

Question

def my_func(num):
   if num ==0:
         
           return 
         
   else:
            print(num)
            my_func(num-1) 
            print("The function is called recursively for",(num),"times")

my_func(7)

結果：

7
6
5
4
3
2
1
The function is called recursively for 1 times
The function is called recursively for 2 times
The function is called recursively for 3 times
The function is called recursively for 4 times
The function is called recursively for 5 times
The function is called recursively for 6 times
The function is called recursively for 7 times

我可以理解打印的數字 1 到 7，但為什么報表中的數字從“1 倍”增加到“7 倍”？ 一旦 num = 0 不是 function “返回”意味着它不會脫離 function 嗎？ 為什么它一直打印語句 1 到 7，為什么數字在增加？ 當它達到 num = 0 時，我會認為它會脫離 function 並且永遠不會到達打印命令？

Answer 1

本質上，這就是您的代碼的工作方式：請注意，縮進用於顯示它正在運行的“功能”。

my_func(7)
 - print 7
 - my_func(6)
   - print 6
   - my_func(5)
     - print 5
     - my_func(4)
       - print 4
       - my_func(3)
         - print 3
         - my_func(2)
           - print 2
           - my_func(1)
             - print 1
             - my_func(0)
               - return
             - print "Called... 1"
           - print "Called... 2"
         - print "Called... 3"
       - print "Called... 4"
     - print "Called... 5"
   - print "Called... 6"
 - print "Called... 7"
Finished everything

但是，在最后的打印語句之間不應該有任何東西，就像 juanpa.arrivillaga 所說的那樣。

Answer 2

遞歸函數是有效嵌套的，因此您的打印行調用print(num)為 7，然后 function 再次以 6 開始，因此您的print(num)再次被調用為 6。

一旦達到 0，就會為每個 function 調用底部語句，因為它們是遞歸完成的。

所以你對於 num = 3，你有效地得到：

my_func(3)

if 3 == 0:
    return 
         
else:
    print(3)

        if 2 == 0:
            return
        
        else:
            print(2)
            
            if 1 == 0:
                return

            else:
                print(1)

                if 0 == 0:
                    return
                
            print("The function is called recursively for",(1),"times")

        print("The function is called recursively for",(2),"times")

    print("The function is called recursively for",(3),"times")

# end of outer-most function

我沒有在每行之間打印出單獨的遞增數字