python中，exit，return,

    有这样一道题目：  字符串标识符.修改例 6-1 的 idcheck.py 脚本,使之可以检测长度为一的标识符,并且可以识别 python 关键字,对后一个要求,你可以使用 keyWord 模块(特别是 keyword.kelist)来帮你.

   我最初的代码是：

#!/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
                    

            
        
        
        
        

    代码完毕后，我测试每一条分支，测试到分支时，必须输入_d4%等包含非法字符的标识符才能进行测试，我最初以为，sys.exit(0)---正常退出脚本，sys.exit(1)非正常退出脚本，但是实际情况是/9sys.exit(1)，仅输出返回码不同）： 

  if (item not in legalstring):
      print "%s isn't legal identifier for Python!" % idInput
     sys.exit(0)

Input your words,please!_d4%
_d4% isn't legal identifier for Python!

Traceback (most recent call last):
  File "E:/python/idcheck.py", line 37, in <module>
    sys.exit(0)
SystemExit: 0
>>> 

    由此可见，这样做没有达到我预期如下输出的效果，那么，问题在哪里呢？在于sys.exit()始终会抛出一个SystemExit异常。

Input your words,please!_d4%
_d4% isn't legal identifier for Python!

#!/usr/bin/env python

import string
import keyword
import sys
import traceback

try:
    #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()
                print "%s is legal identifier for Python!2" % idInput
            else:
                print "%s isn't legal identifier for Python!3" % idInput

except SystemExit:
    pass
except:
    traceback.print_exc()

上面的代码获取sys.exit()抛出的SystemExit异常。

return：在定义函数时从函数中返回一个函数的返回值，终止函数的执行。

exit:下面的代码中，如果把sys.exit()替换成exit，则exit仅仅跳出离它最近的for循环， print "%s is legal identifier for Python!2" % idInput语句会被输出，这里，exit的作用类似于break. 但实际上break和exit作用并不同

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