递归函数中的for循环后，Python变量失去了价值

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这是一个Python bug吗？变量在递归函数的for循环后丢失了值。这是测试代码。我真的在解析XML。

def findversion(aNode, aList, aFlag):
    print "FindVersion ", aNode[0:1]
    print "Findversion ", aFlag
    if aNode[1].find('Software') != -1:
        aFlag = 1
        aList.append(aNode[1])
    if aFlag == 1 and aNode[0] == 'b':
        aList.append(aNode[1])
    print "Before for ", aFlag
    for elem in aNode[2:]:
        print "After for ", aFlag
        findversion(elem,aList,aFlag)
node = ['td', 'Software version']
node2 = ['b', '1.2.3.4' ]
node3 = [ 'td', ' ', node2 ]
node4 = [ 'tr', ' ', node, node3 ]
print node4
myList = list()
myInt = 0
findversion(node4,myList,myInt)
print "Main ",myList
在下面的程序输出中，我总是希望输出前和输出后是相同的。
程序输出。
['tr', ' ', ['td', 'Software version'], ['td', ' ', ['b', '1.2.3.4']]]
FindVersion  ['tr']
Findversion  0
Before for  0
After for  0
FindVersion  ['td']
Findversion  0
Before for  1
After for  0
FindVersion  ['td']
Findversion  0
Before for  0
After for  0
FindVersion  ['b']
Findversion  0
Before for  0
Main  ['Software version']
Python version:
Python 2.7.3 (default, Dec 18 2012, 13:50:09)
[GCC 4.5.3] on cygwin
Type "help", "copyright", "credits" or "license" for more information.


         
          
           python


         
          
           recursion


        
         
          
           
           
            Robert Jacobs
           
          
          
           发布于
           
           2013-03-21


          
           
            
             
             
              Andrew Clark
             
            
            
             发布于
             
             2013-03-21


          
           
            已采纳


          
           
            
             混乱的输出是因为
             
              After for 0
             
             的输出来自函数的不同递归调用（与刚才的
             
              Before for 0
             
             的输出不是同一个）。
            
            
             这里是你的函数的一个版本，有一些额外的信息来跟踪递归调用的深度。
            
            def findversion(aNode, aList, aFlag, i=1):
    print "FindVersion ", aNode[0:1], 'call:', i
    print "Findversion ", aFlag, 'call:', i
    if aNode[1].find('Software') != -1:
        aFlag = 1
        aList.append(aNode[1])
    if aFlag == 1 and aNode[0] == 'b':
        aList.append(aNode[1])
    print "Before for ", aFlag, 'call:', i
    for elem in aNode[2:]:
        print "After for ", aFlag, 'call:', i
        findversion(elem,aList,aFlag,i+1)
这里是新的输出，显示了我正在谈论的内容。
FindVersion  ['tr'] call: 1
Findversion  0 call: 1
Before for  0 call: 1
After for  0 call: 1
FindVersion  ['td'] call: 2
Findversion  0 call: 2
Before for  1 call: 2         # this is from the recursive call
After for  0 call: 1          # this is from the original call
FindVersion  ['td'] call: 2
Findversion  0 call: 2
Before for  0 call: 2
After for  0 call: 2
FindVersion  ['b'] call: 3
Findversion  0 call: 3
Before for  0 call: 3
Main  ['Software version']


          
           
            
             
              
               
               
                Pavel Anossov
               
              
              
               发布于
               
               2013-03-21


          
           
            
             
              
               这个
               
                After
               
               是来自封闭的
               
                findversion
               
               调用。
              
              print Before for 0
start iterating over aNode
   first td:
     print After for  0
     call findversion
       print FindVersion  ['td']
       print Findversion  0
       find Software, set aFlag = 1
       print Before for 1            <---
       start iterating over aNode
       it's empty
   second td:
     print After for  0              <---


          
           
            
             
              
               这不是一个错误。变量aFlag只是特定函数调用的局部，因为它是通过值传递的。当你的程序打印'Before for 1'时，它从未进入for循环，因为aNode[2:]是空的（aNode在那时只有两个元素）。因此，它从不打印任何'After for'，而是立即返回。
              
              
               如果你把打印语句'After for'真正放在for循环之后，而不是放在for循环里面，输出会更清晰。那么输出就会一致。
              
              print "Before for ", aFlag
for elem in aNode[2:]:    
    findversion(elem,aList,aFlag)
print "After for ", aFlag


          
           
            
             
              
               
                
                
                 Robert Jacobs
                
               
               
                发布于
                
                2013-03-21


          
           
            
             
              
               
                所以为了修复我的程序，我需要返回这个标志。
               
               def findversion(aNode, aList, aFlag):
    print "FindVersion ", aNode[0:1]
    print "Findversion ", aFlag
    if aNode[1].find('Software') != -1:
        aFlag = 1
        aList.append(aNode[1])
    if aFlag == 1 and aNode[0] == 'b':
        aList.append(aNode[1])
        aFlag = 0
    print "Before for ", aFlag