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
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
>>>

由此可見，這樣做沒有達到我預期如下輸出的效果，那么，問題在哪里呢？在于sys.exit()始終會拋出一個SystemExit異常。

代碼如下:

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

代碼如下:

#!/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()

上面的代碼獲取sys.exit()拋出的SystemExit異常。

return：在定義函數時從函數中返回一個函數的返回值，終止函數的執行。

exit:下面的代碼中，如果把sys.exit()替換成exit，則exit僅僅跳出離它最近的for循環， print "%s is legal identifier for Python!2" % idInput語句會被輸出，這里，exit的作用類似于break. 但實際上break和exit作用并不同

代碼如下:

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

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