我试图在Python中实现方法重载:
class A:
def stackoverflow(self):
print 'first method'
def stackoverflow(self, i):
print 'second method', i
ob=A()
ob.stackoverflow(2)
但是输出是第二种方法2;类似的:
class A:
def stackoverflow(self):
print 'first method'
def stackoverflow(self, i):
print 'second method', i
ob=A()
ob.stackoverflow()
给了
Traceback (most recent call last):
File "my.py", line 9, in <module>
ob.stackoverflow()
TypeError: stackoverflow() takes exactly 2 arguments (1 given)
我该怎么做呢?
Python在PEP-3124中添加了@overload装饰器,为通过类型检查进行重载提供语法糖——而不仅仅是使用重写。
关于通过PEP-3124中的@overload重载的代码示例
from overloading import overload
from collections import Iterable
def flatten(ob):
"""Flatten an object to its component iterables"""
yield ob
@overload
def flatten(ob: Iterable):
for o in ob:
for ob in flatten(o):
yield ob
@overload
def flatten(ob: basestring):
yield ob
由@overload-decorator转换为:
def flatten(ob):
if isinstance(ob, basestring) or not isinstance(ob, Iterable):
yield ob
else:
for o in ob:
for ob in flatten(o):
yield ob
我用Python 3.2.1写了我的答案。
def overload(*functions):
return lambda *args, **kwargs: functions[len(args)](*args, **kwargs)
工作原理:
Overload接受任意数量的可调用对象,并将它们存储在元组函数中,然后返回lambda。
接受任意数量的参数,
然后返回存储在函数[number_of_unnamed_args_passed]中的调用函数的结果,并将参数传递给lambda。
用法:
class A:
stackoverflow=overload( \
None, \
#there is always a self argument, so this should never get called
lambda self: print('First method'), \
lambda self, i: print('Second method', i) \
)