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python - 我可以在 Python 中为继承的方法创建别名吗?

转载 作者:太空宇宙 更新时间:2023-11-03 15:57:43 25 4
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我的 Python 程序中有一个相当复杂的类层次结构。该程序有很多工具,要么是模拟器,要么是编译器。两种类型共享一些方法,因此有一个 Shared 类作为所有类的基类。精简示例如下所示:

class Shared:
__TOOL__ = None

def _Prepare(self):
print("Preparing {0}".format(self.__TOOL__))


class Compiler(Shared):
def _Prepare(self):
print("do stuff 1")
super()._Prepare()
print("do stuff 2")

def _PrepareCompiler(self):
print("do stuff 3")
self._Prepare()
print("do stuff 4")


class Simulator(Shared):
def _PrepareSimulator(self): # <=== how to create an alias here?
self._Prepare()


class Tool1(Simulator):
__TOOL__ = "Tool1"

def __init__(self):
self._PrepareSimulator()

def _PrepareSimulator(self):
print("do stuff a")
super()._PrepareSimulator()
print("do stuff b")

我可以将方法 Simulator._PrepareSimulator 定义为 Simulator/Shared._Prepare 的别名吗?

我知道我可以创建本地别名,例如:__str__ = __repr__,但在我的情况下,_Prepare 在上下文中未知。我没有 self 也没有 cls 来引用此方法。

我可以编写一个装饰器来返回_Prepare而不是_PrepareSimulator吗?但是我如何在装饰器中找到 _Prepare 呢?

我也需要调整方法绑定(bind)吗?

最佳答案

我设法创建了一个基于装饰器的解决方案,它确保了类型安全。第一个装饰器注释本地方法的别名。它需要保护别名目标。需要第二个装饰器来替换 Alias 实例并检查别名是否指向类型层次结构中的方法。

装饰器/别名定义:

from inspect import getmro

class Alias:
def __init__(self, method):
self.method = method

def __call__(self, func):
return self

def HasAliases(cls):
def _inspect(memberName, target):
for base in getmro(cls):
if target.__name__ in base.__dict__:
if (target is base.__dict__[target.__name__]):
setattr(cls, memberName, target)
return
else:
raise NameError("Alias references a method '{0}', which is not part of the class hierarchy: {1}.".format(
target.__name__, " -> ".join([base.__name__ for base in getmro(cls)])
))

for memberName, alias in cls.__dict__.items():
if isinstance(alias, Alias):
_inspect(memberName, alias.method)

return cls

使用示例:

class Shared:
__TOOL__ = None

def _Prepare(self):
print("Preparing {0}".format(self.__TOOL__))


class Shared2:
__TOOL__ = None

def _Prepare(self):
print("Preparing {0}".format(self.__TOOL__))


class Compiler(Shared):
def _Prepare(self):
print("do stuff 1")
super()._Prepare()
print("do stuff 2")

def _PrepareCompiler(self):
print("do stuff 3")
self._Prepare()
print("do stuff 4")


@HasAliases
class Simulator(Shared):
@Alias(Shared._Prepare)
def _PrepareSimulatorForLinux(self): pass

@Alias(Shared._Prepare)
def _PrepareSimulatorForWindows(self): pass


class Tool1(Simulator):
__TOOL__ = "Tool1"

def __init__(self):
self._PrepareSimulator()

def _PrepareSimulator(self):
print("do stuff a")
super()._PrepareSimulatorForLinux()
print("do stuff b")
super()._PrepareSimulatorForWindows()
print("do stuff c")

t = Tool1()

@Alias(Shared._Prepare) 设置为 @Alias(Shared2._Prepare) 将引发异常:

NameError: Alias references a method '_Prepare', which is not part of the class hierarchy: Simulator -> Shared -> object.

关于python - 我可以在 Python 中为继承的方法创建别名吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40622756/

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