在Python中使用装饰器进行类型检查 [英] Using Decorators in Python for Type Checking
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问题描述
这更多是语法错误问题,我正在尝试在Python Decorators上完成本教程
http://www.learnpython.org/page/Decorators
我的尝试密码
def Type_Check(correct_type):
def new_function(old_function):
def another_newfunction(arg):
if(isintance(arg,correct_type)):
return old_function(arg)
else:
print Bad Type
#此处输入代码
@Type_Check(int)
def Times2(num):
返回num * 2
print Times2(2)
Times2('Not A Number')
@Type_Check(str)
def First_Letter(word):
返回字[0]
打印First_Letter('Hello World' )
First_Letter(['Not','A','String'])
我想知道怎么了,请帮忙
解决方案
似乎您忘了在装饰器的末尾返回新定义的函数:
def Type_Check( correct_type):
def new_function(old_function):
def another_newfunction(arg):
if(isinstance(arg,correct_type)):
return old_function(arg)
否则:
打印错误类型
返回another_newfunction
返回new_function
编辑:还有一些类型,由安德烈(Andrean)修复
This is more of a syntax error issue, I am trying to do this tutorial on Python Decorators
http://www.learnpython.org/page/Decorators
My Attempted Code
def Type_Check(correct_type):
def new_function(old_function):
def another_newfunction(arg):
if(isintance(arg, correct_type)):
return old_function(arg)
else:
print "Bad Type"
#put code here
@Type_Check(int)
def Times2(num):
return num*2
print Times2(2)
Times2('Not A Number')
@Type_Check(str)
def First_Letter(word):
return word[0]
print First_Letter('Hello World')
First_Letter(['Not', 'A', 'String'])
I am wondering whats wrong, please help
解决方案
It looks like you forgot to return the newly defined function at the end of the decorator :
def Type_Check(correct_type):
def new_function(old_function):
def another_newfunction(arg):
if(isinstance(arg, correct_type)):
return old_function(arg)
else:
print "Bad Type"
return another_newfunction
return new_function
EDIT : there was also some types, fixed by andrean
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