<p>关于继承问题:这可以通过对MultiMethod稍作修改来实现。(迭代self.typemap并用<code>issubclass</code>检查):</p>
<pre><code>registry = {}
class MultiMethod(object):
def __init__(self, name):
self.name = name
self.typemap = {}
def __call__(self, *args):
types = tuple(arg.__class__ for arg in args) # a generator expression!
for typemap_types in self.typemap:
if all(issubclass(arg_type,known_type)
for arg_type,known_type in zip(types,typemap_types)):
function = self.typemap.get(typemap_types)
return function(*args)
raise TypeError("no match")
def register(self, types, function):
if types in self.typemap:
raise TypeError("duplicate registration")
self.typemap[types] = function
def multimethod(*types):
def register(function):
name = function.__name__
mm = registry.get(name)
if mm is None:
mm = registry[name] = MultiMethod(name)
mm.register(types, function)
return mm
return register
class A(object):
pass
class B(A):
pass
class C(object):
pass
@multimethod(A,A)
def foo(arg1,arg2):
print 'works'
foo(A(),A()) #works
foo(A(),B()) #works
foo(C(),B()) #raises TypeError
</code></pre>
<p>注意,<code>self.typemap</code>是dict,dict是无序的。因此,如果使用@multimethod注册两个函数,其中一个函数的类型是另一个函数的子类,那么{<cd3>}的行为可能是未定义的。也就是说,结果将取决于在循环<code>for typemap_types in self.typemap</code>中首先出现的<code>typemap_types</code>。在</p>