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：

复制代码 代码如下:
#!/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）：

复制代码 代码如下:
  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
    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。

复制代码 代码如下:
Input your words,please!_d4%
_d4% isn\’t legal identifier for Python!

复制代码 代码如下:
#!/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

复制代码 代码如下:
                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