Python assignment operator differs from non assignment

问题

I have face this weird behavior I can not find explications about.

MWE:

l = [1]
l += {'a': 2}
l
[1, 'a']
l + {'B': 3}
Traceback (most recent call last):
  File "<input>", line 1, in <module>
TypeError: can only concatenate list (not "dict") to list

Basically, when I += python does not raise an error and append the key to the list while when I only compute the + I get the expected TypeError.

Note: this is Python 3.6.10

回答1:

l += ... is actually calling object.__iadd__(self, other) and modifies the object in-place when l is mutable

The reason (as @DeepSpace explains in his comment) is that when you do l += {'a': 2} the operation updates l in place only and only if l is mutable. On the other hand, the operation l + {'a': 2} is not done in place resulting into list + dictionary -> TypeError.

(see here)

l = [1]
l = l.__iadd__({'a': 2})
l
#[1, 'a']

is not the same as + that calls object.__add__(self, other)

l + {'B': 3}

TypeError: can only concatenate list (not "dict") to list

回答2:

So as the authors say this is not a bug. When you Do a += b it is like b come to a's house and changing it the way that a like it to be. what the Authors say is when you do a + b it cannot be decided which one's style will get prioritized. and no one knows where will the result of a + b will go until you execute it. So you can't decide whose style it would be. if it is a style it would be [1, 'a']'s, and if it is b style it would be an error. and therefore it cannot be decided who will get the priority. So I don't personally agree with that statement. because when you take the call stack a is in a higher place than b. when there is a expression like a + b you first call a.__add__(self, other) if a.__add__ is NotImplemented (in this case it is implemented). then you call a.__radd__(self, other). which means call other.__add__ in this case b.__add__. I am telling this based on the place of the call stack and the python community may have more important reasons to do this.

来源：https://stackoverflow.com/questions/61457122/python-assignment-operator-differs-from-non-assignment

标签

python

assignment-operator