Ralf Hemmecke <[EMAIL PROTECTED]> writes:
> On 01/10/2008 09:42 AM, Martin Rubey wrote:
> > Dear Gaby,
> > I saw you implemented map!$Stack as follows:
> > map!(f: S -> S, s: %) == -- from HOAGG
> > map!(f, deref s)$List(S)
> > s
> > I have two questions: 1) why didn't you say
> > map!(f: S -> S, s: %) == ref map!(f, deref s)$List(S)
> > 2) isn't map!(f: S -> S, s: %) == -- from HOAGG
> > map!(f, deref s)$List(S)
> > s
> > assuming that map! stores its result in s? I'm not sure whether this is
> > ok:
>
> Unfortunately 1) and 2) are not equivalent with the current implementation of
> the Reference domain. In boolean.spad.pamphlet one finds
>
> Reference(S:Type): Type with
> ref : S -> %
> ++ ref(n) creates a pointer (reference) to the object n.
> ...
> == add
> Rep := Record(value: S)
> p = q == EQ(p, q)$Lisp
> ref v == [v]
> ...
>
> That looks to me like that for
>
> a: Reference Integer := ref 1;
> b: Reference Integer := ref deref a;
>
> it does not hold
>
> a = b
Quite possible. But I wouldn't need that in map! anyway, would I?
the doc to map! only says
map!(f,u) destructively replaces each element x of u by \axiom{f(x)}.
and I think that should be interpreted as:
the result of map!(f,u) is obtained by replacing each element x of u by
f(x). In this process, u may be destroyed.
I think that would allow the implementation
map!(f: S -> S, s: %) == ref map!(f, deref s)$List(S)
but *not*
map!(f: S -> S, s: %) == -- from HOAGG
map!(f, deref s)$List(S)
s
> That said ...
>
> *Don't use l after you have called sort!(l) unless you say l:=sort!(l).*
That's exactly what I'm thinking. (for sort! it is obvious, but not that much
for map!, that's why I asked)
Martin
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