super in Python 3 and variadic arguments
Ned Batchelder
ned at nedbatchelder.com
Wed Oct 9 12:47:13 EDT 2013
On 10/9/13 11:44 AM, Marco Buttu wrote:
> Given this class:
>
> >>> class A:
> ... def afoo(*args):
> ... print(args)
>
> in Python 3 we can write the following class:
>
> >>> class B(A):
> ... def bfoo(*args):
> ... super(B, args[0]).afoo(*args[1:])
> ...
> >>> B().bfoo(1, 2, 3)
> (<__main__.B object at 0x7f5b3bde48d0>, 1, 2, 3)
>
>
> without giving arguments to super, in this way:
>
> >>> class B(A):
> ... def bfoo(self, *args):
> ... super().afoo(*args)
> ...
> >>> B().bfoo(1, 2, 3)
> (<__main__.B object at 0x7f5b3bdea0d0>, 1, 2, 3)
>
> But it does not work in this case:
>
> >>> class B(A):
> ... def bfoo(*args):
> ... super().afoo(*args[1:])
> ...
> >>> B().bfoo(1, 2, 3)
> Traceback (most recent call last):
> File "<stdin>", line 1, in <module>
> File "<stdin>", line 3, in bfoo
> RuntimeError: super(): no arguments
>
> How come?
The no-args super() call inspects the calling environment to determine
the class and self. "self" is the first local name stored in
frame.f_code.co_localsplus, but *args doesn't put "args" into that entry
of the code object. Basically, super() is looking for the first regular
argument in the function.
--Ned.
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