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 <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。

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