"inherited" keyword in Python?

Paul Foley see at below
Wed Dec 6 02:30:35 CET 2000

On Tue, 5 Dec 2000 22:11:08 +0100, Alex Martelli wrote:

> To clarify: Python's rule (as applied by, e.g., getattr) is anything
> but broken, but it doesn't easily let you return (for a given class X

Python's rule works for Python, sure -- but Python doesn't provide any
way to call a superclass method without knowing exactly what you want
to call (else all this would be unnecessary!)

> that inherits from A, B, C, ...) a single class Z that will ensure that
> getattr(Z,whatever) is the same as getattr(X,whatever) for any
> whatever that is not a key in X.__dict__.

I'm not sure that's all that's needed, though.

>> that won't work.  I thought super() was just supposed to call the
>> appropriate same-named method of a parent class.  Anything else can be
>> done through the usual self.whatever, surely?

> Not really.  To translate literally a Java class which has methods X
> (which calls the base's Y) and Y (which calls the base's X method),
> I'd need a Python's super() to be like Java's super (returning a class

I've never bothered with Java.  How can you write a method X which
calls the base's Y without explicitly calling the base's Y?  And if
you have to do it explicitly, that'll work in Python, too.

> object reference) and not like what I name 'supercall'.  self.X and
> self.Y just won't do for this.

It seems to me, what you said above about Java notwithstanding, that
the only time it's OK to call X where X is a method in a base class,
and where self.__class__ also has a method named X (so that self.X()
will do the wrong thing), is from inside self.X.

>> > But, how often is it going to be satisfactory to call 'just
>> > the one' (of several) inherited-method of a certain name, in
>> > a multiple-inheritance context?  Just as often we may want
>> It isn't (at least, not to me), which is why I avoided doing it that

> But you did do that -- you call just one of the several 'inherited'
> implementations of the method, in the code below.

Any one call to super() only calls one method, naturally.  But if
everyone calls super() _all_ the methods eventually get called.  Under
"Python rules", only the methods in D, B and A can ever run -- the one
in C never does, unless called explicitly with C.foo(self) from D; and
in that case, it's called out of order -- you either get D, B, A, C or
D, C, B, A, or, more likely D, B, A, C, A or D, C, A, B, A, with A.foo
getting called twice, since both B and C inherit A.  This sucks.  I
get D, B, C, A, which is the right order (B.foo and C.foo are both
called before their common superclass's A.foo, B.foo is called before
C.foo since B is to the left of C in D.__bases__) and A.foo is only
called once.

>> way.  Having figured out how to get the frame info, here's an
>> implementation.  No need to pass in the object and method name:

> I'd still want to pass them, at least optionally, to handle the
> above case where I want to call a _different_ inherited method.

I'm working on the assumption that if you want to do that, you're
doing something wrong.

>> If you just call super(__thisclass) with no args other than the class,
>> it passes on the args the calling method got; or you can specify other
>> args explicitly [yes, I know this prevents you calling with no args if
>> super's caller has args]

> As, once again, convenience gets in the way of simplicity and

Maybe.  This is an attempt to copy the CALL-NEXT-METHOD functionality
of CLOS; in CLOS, all the methods will have similar argument lists,
because they're all part of one generic function, not independent
members of various classes.  Just chop out the 5 lines in the middle
that copy the caller's arglist if you think it's better not to do

> generality.  Oh well:-).  "Tantum convenientia potuit suadere
> malorum", as Lucretius might write if he lived today!-)

[Am I supposed to be one of them?  And shouldn't "malorum" be dative?]

Cogito ergo I'm right and you're wrong.                 -- Blair Houghton

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