Python 空字典没有通过引用传递？

Question

这个示例代码有点奇怪，但请耐心等待...

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': {}}

我本以为我的 {'123' = 'abc'} 已经完成了第二次打印，因为字典应该已经传递到递归 __init__通过引用调用。但是，如果我更改此行:

self._internal_dict[attribute_name] = {}

为此:

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

然后我得到以下打印:

{'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？

Answer 1

问题是:

self._internal_dict = internal_dict or {}

一个空的字典是假的，所以你会在后续的递归调用中得到一个新的空字典。这就是为什么将 dict 初始化为非空(真实)“修复”它的原因。

你想要:

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