Addressing the last element of a list

[Antoon Pardon]

Op 2005-11-15, Mike Meyer schreef <mwm at mired.org>:
> Antoon Pardon <apardon at forel.vub.ac.be> writes:
>>>> Like having an assignment operator (let use @= for it) next to a
>>>> (re)bind operator.
>>>> We could then have something like the following.
>>>> a = 5
>>>> b = a
>>>> a @= 7
>>>> b ==> would result in 7.
>>> You've just overwritten the object referred to by the token "5" in the
>>> source code with the value 7, so you get:
>>> print 5
>>> 7
>> You have a valid point, but I think a different approach is possible.
>> Don't make the distinction between mutable and immutable types but
>> between mutable and immutable objects. So an int wouldn't be an
>> immutable type, but 5 would be an immutable object.
>> So the code above I gave would throw an exception, but the following
>> might work.
>> a @= 5
>> b = a
>> b @= 7
>> a ==> would result in 7.
>
> Which solves that issue but brings up - well, more issues. What
> happens if the third line is 'b @= "seven"'? Does this require an
> extra level of indirection in the implementation? Avoiding that kind
> of thing was the reason for suggesting this:

It depends on how far you want to go.

I think one can argue that in case of an inplace replacement, this
only makes sense if the two objects belong to the same class. So
no extra level of indirection is then required.

Otherwise it really depends on how the builtin objects are implemented.
For user classes, a somewhat simplistic implementation could be
something like:

  def '@=' (one, two):
    one.__dict__.clear()
    one.__dict__.update(two.__dict__)

But if such an approach is usable with the builtin classes, I don't
know.  I'm sure this will have it's own issues though.

>>> Another approach is to have a special object type if you want to allow
>>> assignment to the object it refers to. That's what Python does now, it
>>> just doesn't have a syntax just to support that. If it did, your
>>> example might look like:
>>> a := 5
>>> b = @a
>>> a := 7
>>> @b ==> would result in 7
>> Could it also work the other way? By somehow manipulating b change a?
>
> It's intended to work the other way as well. But the second line is
> wrong: it should have been "b = a". "b = @a" assigns b the value
> referenced by a, and @b would then be an error. "b = a" assigns b to
> the reference that is a, so that @b returns it's value.
>
> The critical thing is that this doesn't introduce any new facilities
> into the language, just some new syntax, so there's no implementation
> impact. In fact, if you're willing to put up with some notational
> abuse, we can do this now:
>
> class Ref(object):
>     _unbound = object()
>     def __new__(cls, value = _unbound):
>         """We're an object, but need to ignore the optional argument."""
>
>         return object.__new__(cls)
>
>     def __init__(self, value = _unbound):
>         """Bind the  optional value, if provided."""
>         if value is not self._unbound:
>             self._value = value
>
>     def __pos__(self):
>         """Return value, if bound."""
>         try:
>             return self._value
>         except AttributeError:
>             raise ValueError, "%s object does not have a value stored." % \
>                   self.__class__.__name__
>
>     def __iadd__(self, value):
>         self._value = value
>         return self
>
> Usage:
>
>>>> x = Ref.Ref()
>>>> x += 23
>>>> +x
> 23
>>>> a = x
>>>> x += 25
>>>> +a
> 25
>>>> a += "this is a test"
>>>> +x
> 'this is a test'
>
> Since it doesn't have real lannguage support, things like +x += 25
> don't work. That's the only obvious gotcha. Maybe ~ would be a better
> prefix.

I'm a bit puzzled on how you would implement this as real language
support. As far as I understand this only works with Ref objects.

You can't do something like the following.

l = [3, 7]
a = Ref(l[0])

Instead you would have to do something like
l = [Ref(3), Ref(7)]
a = l[0]

So it seems that if you want to use this idea for implementing langauge
support you will have to wrap all objects in a Ref internally and this
seems some kind of extra indirection too.

-- 
Antoon Pardon