Python空字典没有通过引用传递? [英] Python empty dict not being passed by reference?
问题描述
这个示例代码有点奇怪,但请耐心等待...
This sample is code is a bit odd, but bear with me...
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'}
已经通过第二次打印,因为字典应该已经传递到递归 __init__
参考调用.但是,如果我更改这一行:
I would have expected my {'123' = 'abc'}
to have gotten through to the second print because the dictionary should have been passed into the recursive __init__
call by reference. However, if I change this line:
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'}}
为什么启动该字典数据会导致它通过引用正确传递?
Why does priming that dictionary data cause it to be passed properly by reference?
推荐答案
这是问题所在:
self._internal_dict = internal_dict or {}
空字典是假的,因此您将在后续递归调用中得到一个 新 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.
An empty dict is falsy, so you will get a new empty dict on subsequent recursive calls. That's why initializing the dict to non-empty (truthy) "fixes" it.
你想要:
self._internal_dict = {} if internal_dict is None else internal_dict
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