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From: Vladimir Prus (ghost_at_[hidden])
Date: 2004-01-30 08:31:06

David Abrahams wrote:
> Vladimir Prus <ghost_at_[hidden]> writes:
> > There are two alternatives. First is not sorting them anything all. This
> > might cause too many property-sets which are equvivalent but have
> > different property order. That can slow things down.
> FWIW if you don't care about repetitions, you can quickly tell
> whether two lists have the same elements by:
> if $(l1) in $(l2) && $(l2) in $(l1)

I don't think that will help. The issue is that if we decide that

<foo>xxx <foo>yyy


<foo>yyy <foo>xxx

are different for all 'foo', then we have to create much more property-set
instances that we do now. If we decide they are difference only of 'foo' is
'order-sensitive' feature, we have to create more property sets, but not as
much. The question is if there's real performance effect.

- Volodya


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