Question

这是一个python bug吗？在递归函数中for循环后变量丢失值。这是测试代码。我正在解析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，我总是希望输出前输出与后输出相同。

节目输出

['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版

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.

Answer 1

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（与其上方的Before for 0输出不同）。

以下是函数的一个版本，其中包含一些跟踪递归调用深度的额外信息：

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']

Answer 2

此After来自随附的findversion来电：

...
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              <---
     ...

Answer 3

这不是一个错误。变量aFlag仅对特定函数调用是本地的，因为它是按值传递的。当你的程序打印'Before for 1'时它永远不会进入for循环，因为aNode [2：]为空（aNode当时只有两个元素）。因此，它从不打印任何'After for'但立即返回。

如果你把for语句'After for'实际放在for循环之后而不是for循环中，那么输出会更清晰。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

Answer 4

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

Python变量在递归函数中的for循环后丢失值

4 个答案: