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社区首页 >问答首页 >这个策略模式的python实现应该在init方法中使用私有变量吗?

这个策略模式的python实现应该在init方法中使用私有变量吗?
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Stack Overflow用户
提问于 2020-10-07 12:12:00
回答 1查看 281关注 0票数 1

我正在学习策略设计模式,以及Python中的属性装饰器。我遇到了这个例子:

代码语言:javascript
复制
from __future__ import annotations
from abc import ABC, abstractmethod
from typing import List


class Context():
    """
    The Context defines the interface of interest to clients.
    """

    def __init__(self, strategy: Strategy) -> None:
        """
        Usually, the Context accepts a strategy through the constructor, but
        also provides a setter to change it at runtime.
        """

        self._strategy = strategy

    @property
    def strategy(self) -> Strategy:
        """
        The Context maintains a reference to one of the Strategy objects. The
        Context does not know the concrete class of a strategy. It should work
        with all strategies via the Strategy interface.
        """

        return self._strategy

    @strategy.setter
    def strategy(self, strategy: Strategy) -> None:
        """
        Usually, the Context allows replacing a Strategy object at runtime.
        """

        self._strategy = strategy

    def do_some_business_logic(self) -> None:
        """
        The Context delegates some work to the Strategy object instead of
        implementing multiple versions of the algorithm on its own.
        """

        # ...

        print("Context: Sorting data using the strategy (not sure how it'll do it)")
        result = self._strategy.do_algorithm(["a", "b", "c", "d", "e"])
        print(",".join(result))

        # ...


class Strategy(ABC):
    """
    The Strategy interface declares operations common to all supported versions
    of some algorithm.

    The Context uses this interface to call the algorithm defined by Concrete
    Strategies.
    """

    @abstractmethod
    def do_algorithm(self, data: List):
        pass


"""
Concrete Strategies implement the algorithm while following the base Strategy
interface. The interface makes them interchangeable in the Context.
"""


class ConcreteStrategyA(Strategy):
    def do_algorithm(self, data: List) -> List:
        return sorted(data)


class ConcreteStrategyB(Strategy):
    def do_algorithm(self, data: List) -> List:
        return reversed(sorted(data))


if __name__ == "__main__":
    # The client code picks a concrete strategy and passes it to the context.
    # The client should be aware of the differences between strategies in order
    # to make the right choice.

    context = Context(ConcreteStrategyA())
    print("Client: Strategy is set to normal sorting.")
    context.do_some_business_logic()
    print()

    print("Client: Strategy is set to reverse sorting.")
    context.strategy = ConcreteStrategyB()
    context.do_some_business_logic()

我对属性方法/装饰器的理解是,在这种情况下,@property提供了一个接口来设置属性(温度)。在__init__方法中将属性命名为_strategy意味着它应该是一个私有变量。这是错误的还是多余的?我认为这个变量应该命名为strategy,但它的接口应该使用私有变量实现(即在getter/setter中)

摘自https://refactoring.guru/design-patterns/strategy/python/example

编辑:

为了澄清我的推理:是否可以在运行时更改策略,如下所示:

代码语言:javascript
复制
a = Context()
a.strategy = somestrategy
EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2020-10-07 12:26:22

该属性没有命名为_strategy,而是命名为strategy,但是保存其值的内部变量命名为_strategy

这一点:

代码语言:javascript
复制
    def __init__(self, strategy: Strategy) -> None:
        self._strategy = strategy

允许:

代码语言:javascript
复制
c = Context(some_strategy)

但是,当您尝试c._strategy时,将其命名为_strategy会导致警告。

还有这个:

代码语言:javascript
复制
    @property
    def strategy(self) -> Strategy:
        return self._strategy

然后允许:

代码语言:javascript
复制
my_strategy = c.strategy

当访问该属性时,getter返回self._strategy的值。

最后是这个:

代码语言:javascript
复制
     @strategy.setter
     def strategy(self, strategy: Strategy) -> None:
        self._strategy = strategy

允许:

代码语言:javascript
复制
c.strategy = another_strategy

使strategy不仅是一个只读属性,而且是一个读/写属性。

注意:下面的代码没有错,但它做了一些其他的事情:

代码语言:javascript
复制
class Complex:
    def __init__(self, strategy: int) -> None:
        self.strategy = strategy

    @property
    def strategy(self) -> int:
        return self._strategy

    @strategy.setter
    def strategy(self, strategy: int) -> None:
        self._strategy = strategy


c = Complex(1)
print(c.strategy)

不同之处在于,构造函数(__init__)现在不直接设置隐藏属性,而是自己调用属性的设置器。

如果现在有人覆盖了这个类,它仍然会使用它,比较一下:

代码语言:javascript
复制
class MyClass:
    def __init__(self, a, b: int) -> None:
        self.a = a
        self._b = b

    @property
    def a(self) -> int:
        return self._a

    @a.setter
    def a(self, a: int) -> None:
        self._a = a

    @property
    def b(self) -> int:
        return self._b

    @b.setter
    def b(self, b: int) -> None:
        self._b = b


class MySubClass(MyClass):
    @MyClass.a.setter
    def a(self, a: int) -> None:
        self._a = a + 10

    @MyClass.b.setter
    def b(self, b: int) -> None:
        self._b = b + 10


c = MyClass(1, 2)
print(c.a)
print(c.b)

s = MySubClass(1, 2)
print(s.a)
print(s.b)

结果:

代码语言:javascript
复制
1
2
11
2

因此,这取决于您是否希望继承您的类的人能够更改该行为。

票数 2
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/64237084

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