Python assignment operator differs from non assignment

Question

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 ca         


        

Answer 1

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.