Robert Bradshaw wrote:
> On Sep 27, 2007, at 4:43 AM, Jason Grout wrote:
>
>> A=A.union(B)
>>
>> to modify A?
>
> Yes. Or even
>
> A.inplace_union(B)
>
> (or some better name than that) which returns nothing. Otherwise, for
> example, every function that received a graph as an argument would
> have to make a copy to avoid side effects (or the calling function
> would always have to make a copy before passing in just in case).
> Explicitly making a copy everywhere seems cumbersome.
Thanks for the direct and good advice. I guess I had convinced myself
that an object method would naturally do something to the object, so it
was pretty natural to always modify in-place.
>
>> It's slightly worse performance-wise, it seems, because we
>> are always making copies and throwing old copies away. Especially
>> if we
>> have the above (A.union(B).union(C).union(D).union(E))---then we
>> make 4
>> new copies of things and throw away all but 1 of them.
>
> True, this is the price to pay. It might be worth having a function
> that does things inplace if one needs the speed, but I like to avoid
> side effects in general. Though I think the syntax above is the
> clearest, A.union([B,C,D]) could be done more efficiently.
Is this the sort of situation your ref-counting stuff would enhance?
For example, in the above situation, if a new object were returned by
union, then A.union(B) would not be referenced anywhere else in the
system, so A.union(B).union(C) would actually not create a new object,
but do the union in-place in A.union(B)? So the system would understand
that the following code:
def union(self, othergraph):
g=self.copy()
# Modify g to be the union of g and othergraph
...
return g
should make a copy of self in the A.union(B) case, but not make a copy
when invoked in the .union(C) case above.
Am I understanding the idea behind your ref-counting thread mentioned
elsewhere?
Thanks,
Jason
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