Python空dict未通过引用传递？

这个例子是代码有点奇怪，但请容忍我

class Foo(object):
    def __init__(self, internal_dict = None):
        self._internal_dict = internal_dict or {}

        for attribute_name in self.__class__.__dict__.keys():
            attr = getattr(self.__class__, attribute_name)
            if isinstance(attr, str) and attribute_name.startswith("a"):
                # We are iterating over all string attributes of this class whos name begins with "a" 
                self._internal_dict[attribute_name] = {}
                setattr(self, attribute_name + '_nested_object', Foo(internal_dict=self._internal_dict[attribute_name]))

class FooChild(Foo):
    ax = "5"
    ay = "10"

fc = FooChild()

print fc.ax_nested_object._internal_dict # This prints {}

fc.ax_nested_object._internal_dict['123'] = 'abc'

print fc._internal_dict # This prints {'ay': {}, 'ax': {}}

我本来希望我的

{'123'='abc'}

能够通过第二次打印，因为字典应该通过引用传递到递归的

\uuuuu init\uuu

调用中。但是，如果我更改此行：

self._internal_dict[attribute_name] = {}

为此：

self._internal_dict[attribute_name] = {'test': 1}

然后我会打印以下内容：

{'test': 1}
{'ay': {'test': 1}, 'ax': {'test': 1, '123': 'abc'}}

为什么启动字典数据会导致它通过引用正确传递？

这就是问题所在：

self._internal_dict = internal_dict or {}

空的dict是错误的，因此在后续递归调用中您将得到一个新的空dict。这就是为什么将dict初始化为非空（truthy）“修复”它

你想要：

self._internal_dict = {} if internal_dict is None else internal_dict

这就是问题所在：

self._internal_dict = internal_dict or {}

空的dict是错误的，因此在后续递归调用中您将得到一个新的空dict。这就是为什么将dict初始化为非空（truthy）“修复”它

你想要：

self._internal_dict = {} if internal_dict is None else internal_dict

---

你的类的问题是你在分配

self.\u internal\u dict

变量时所走的捷径

不幸的是，空字典的真值为false

如果将代码更改为：

class Foo(object):
    def __init__(self, internal_dict = None):
        if internal_dict is None:
            internal_dict = {}

        self._internal_dict = internal_dict

        for attribute_name in self.__class__.__dict__.keys():
            attr = getattr(self.__class__, attribute_name)
            if isinstance(attr, str) and attribute_name.startswith("a"):
                # We are iterating over all string attributes of this class whos name begins with "a" 
                self._internal_dict[attribute_name] = {}
                setattr(self, attribute_name + '_nested_object', Foo(internal_dict=self._internal_dict[attribute_name]))

class FooChild(Foo):
    ax = "5"
    ay = "10"

fc = FooChild()

print fc.ax_nested_object._internal_dict # This prints {}

fc.ax_nested_object._internal_dict['123'] = 'abc'

print fc._internal_dict # This prints {'ay': {}, 'ax': {}}

---

你的类的问题是你在分配

self.\u internal\u dict

变量时所走的捷径

不幸的是，空字典的真值为false

如果将代码更改为：

class Foo(object):
    def __init__(self, internal_dict = None):
        if internal_dict is None:
            internal_dict = {}

        self._internal_dict = internal_dict

        for attribute_name in self.__class__.__dict__.keys():
            attr = getattr(self.__class__, attribute_name)
            if isinstance(attr, str) and attribute_name.startswith("a"):
                # We are iterating over all string attributes of this class whos name begins with "a" 
                self._internal_dict[attribute_name] = {}
                setattr(self, attribute_name + '_nested_object', Foo(internal_dict=self._internal_dict[attribute_name]))

class FooChild(Foo):
    ax = "5"
    ay = "10"

fc = FooChild()

print fc.ax_nested_object._internal_dict # This prints {}

fc.ax_nested_object._internal_dict['123'] = 'abc'

print fc._internal_dict # This prints {'ay': {}, 'ax': {}}

---

这就是我在检查是否提供了可选参数时询问是否包含

是否为None

的确切原因。感谢您确认这确实发生了；-）这就是我在检查是否提供了可选参数时询问是否包含

是否为None

的确切原因。感谢您确认这确实发生了；-）谢谢我以为我疯了。谢谢！我以为我疯了。