有這樣一道題目：  字符串標識符.修改例 6-1 的 idcheck.py 腳本,使之可以檢測長度為一的標識符,并且可以識別 Python 關鍵字,對后一個要求,你可以使用 keyword 模塊(特別是 keyword.kelist)來幫你.

我最初的代碼是：

代碼如下:
#!/usr/bin/env python

import string
import keyword
import sys

#Get all keyword for python
#keyword.kwlist
#['and', 'as', 'assert', 'break', ...]
keyWords = keyword.kwlist

#Get all character for identifier
#string.letters ==> 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'
#string.digits  ==> '0123456789'
charForId = string.letters + "_"
numForId = string.digits

idInput = raw_input("Input your words,please!")

if idInput in keyWords:
    print "%s is keyword fot Python!" % idInput
else:
    lenNum = len(idInput)
    if(1 == lenNum):
        if(idInput in charForId and idInput != "_"):
            print "%s is legal identifier for Python!" % idInput
        else:
            #It's just "_"
            print "%s isn't legal identifier for Python!" % idInput

    else:
        if(idInput[0:1] in charForId):
            legalstring = charForId + numForId
            for item in idInput[1:]:
                if (item not in legalstring):
                    print "%s isn't legal identifier for Python!" % idInput
                    sys.exit(0)
            print "%s is legal identifier for Python!2" % idInput
        else:
            print "%s isn't legal identifier for Python!3" % idInput
    

代碼完畢后，我測試每一條分支，測試到分支時，必須輸入_d4%等包含非法字符的標識符才能進行測試，我最初以為，sys.exit(0)---正常退出腳本，sys.exit(1)非正常退出腳本，但是實際情況是/9sys.exit(1)，僅輸出返回碼不同）：
代碼如下:
  if (item not in legalstring):
      print "%s isn't legal identifier for Python!" % idInput