On Sun, Nov 22, 2020 at 07:28:00PM -0000, Jonatan wrote:
> somehow, The expression "super() <OPERATOR> other" does not turn into
> "super().__OPERATOR__(other)".
Is there something documented that lead you to believe that it should?
By the way, for most operators, there is no simple relationship between
the operator and the dunder method called. For example, consider
addition `x + y`. The interpreter does something *roughly* like this:
if y is a subclass of x:
try calling y.__radd__(x)
if it doesn't exist, or if it returns NotImplemented,
try calling x.__add__(y)
if it doesn't exist, or returns NotImplemented,
raise TypeError
otherwise:
try calling x.__add__(y)
if it doesn't exist, or if it returns NotImplemented,
try calling y.__radd__(x)
if it doesn't exist, or returns NotImplemented,
raise TypeError
only I expect I have the details wrong and it may be even more
complicated. The point is, `super() operator other` will not, in
general, turn into a single dunder method, but potentially two or maybe
even more.
> this bug is for any operator that you apply on super().
> I'd be happy to hear from you why it happens and whether will it be fixed in
> python3.10.
Before we can "fix" it, we have to agree that it is a bug and that it
should be fixed. What gives you the impression that this is a bug?
Is the behaviour documented somewhere as working?
If we made this change, would it extend to other dunder methods like
`__str__`, `__len__`, `__enter__`, etc?
--
Steve
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