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这是一个 Python 错误吗？递归函数中的 for 循环后变量丢失值。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，我总是希望输出之前的输出与输出之后的相同。

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']

蟒蛇版本：

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.
4

4 回答 4

0

令人困惑的输出是因为After for 0输出来自函数的不同递归调用（与Before for 0它上面的输出不同）。

这是您的函数的一个版本，其中包含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)

这是新的输出，它显示了我在说什么：

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']
于 2013-03-20T19:33:31.900 回答
0

这After是来自封闭findversion电话：

...
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              <---
     ...
于 2013-03-20T19:33:34.257 回答
0

这不是一个错误。变量 aFlag 仅对特定函数调用是局部的，因为它是按值传递的。当您的程序打印“Before for 1”时，它永远不会进入 for 循环，因为 aNode[2:] 是空的（aNode 当时只有两个元素）。因此，它从不打印任何“After for”，而是立即返回。

如果您将 print 语句 'After for' 实际上放在 for 循环之后而不是 for 循环内，则输出将更加清晰。然后输出将是一致的。

print "Before for ", aFlag
for elem in aNode[2:]:    
    findversion(elem,aList,aFlag)
print "After for ", aFlag
于 2013-03-20T19:37:42.270 回答
0

所以为了修复我的程序，我需要返回标志。

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
    for elem in aNode[2:]:
        print "After for ", aFlag
        aFlag = findversion(elem,aList,aFlag)
    return aFlag
于 2013-03-20T19:46:35.900 回答