On Thu, 10 Jan 2008, Martin Rubey wrote:
|
| 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)
I'm not sure that would be correct; if s and t points to the same
memory cell before call to map!, the should point to the same memory
cell after call to map!. No?
-- Gaby
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