我有旧的类层次结构,我想用具有不同方法签名的新类扩展:覆盖的方法应返回树值,而不是两个。旧有类的第三个返回值应该是一些默认值。下面的示例(repl.it)。
class BaseClass:
def do_work(self):
param1, param2, param3 = self.get_params()
self._process(param1, param2, param3)
def get_params(self):
# this method shoud be overrided in the child class
raise NotImplementedError()
def _process(self, param1, param2, param3):
# do some process, for example, print params
print(param1, param2, param3)
class LegacyChild(BaseClass):
def get_params(self):
return 1, 2, 3
class NewChild(BaseClass):
def get_params(self):
return 1, 2
legacy_child = LegacyChild()
legacy_child.do_work()
new_child = NewChild()
new_child.do_work()
在这种情况下,我在第三行得到ValueError: not enough values to unpack (expected 3, got 2)
。这是正确的。
我解决这个问题的第一个想法是将硬编码的三个参数更改为可变长度的元组,然后根据其长度设置param1
,param2
和param3
:
class BaseClass:
def do_work(self):
params = self.get_params()
if len(params) == 3:
param1, param2, param3 = params
else:
param1, param2 = params
param3 = 3 # some default value
self._process(param1, param2, param3)
def get_params(self):
# this method shoud be overrided in the child class
raise NotImplementedError()
def _process(self, param1, param2, param3):
# do some process, for example, print params
print(param1, param2, param3)
我认为此解决方案过于粗糙(repl.it)。
该问题还有其他可能的解决方法吗?
也许这吗?
param1, param2, *param3 = self.get_params() # param3 is an array. When there are only 2 elements, it's empty.
self._process(param1, param2, (param3 or [3])[0]) # 3 is the default value
这是假设get_params
将始终产生2或3个元素。无法轻易扩展它以处理仅返回一个元素的方法。