Linear component may be abandoned.

[Steinway Wu]

Hi, 

In `stream_vt.dats`, I saw the following snippets,

fun {a:vt0p} {b:vt0p} stream_vt_map_con (xs: stream_vt (a)) : stream_vt_con (b) = let

    val xs_con = !xs

in

    case+ xs_con of

    | @stream_vt_cons(x, xs) => let

        val y = stream_vt_map$fopr<a><b> (x) // fopr is of type {a,b:vt@ype} &a >> a?! -> b

        val xs = xs

        val () = free@ (xs_con)

      in

        stream_vt_cons{b}(y, stream_vt_map<a><b> (xs))

      end 

    | ~stream_vt_nil((*void*)) => stream_vt_nil()

end

and I have three questions. 

1. what is a?! 

2. why fopr takes &a, instead of just a?

3. what happened when val xs = xs 

Thanks, 

[Hongwei Xi]

1.

When [a] is a linear type, a?! is only the data part of [a] (that is, there is no view in a?!).

When [a] is non-linear, then a?! is just [a].

2. When [a] is large (containing many bytes), call-by-reference can be more efficient. The convention

    in ATS is to use call-by-reference when linear data is handled.

3. The 'xs' on the right-hand side is a variable (with address); its content needs to be taken out before free@ can be called.

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[Steinway Wu]

What about this? Let the worker function consume a. And in map, free the constructor, instead of unfold the constructor. 

fun {a,b:vt@ype} stream_vt_map_con (xs: stream_vt a): stream_vt_con b
fun {a,b:vt@ype} stream_vt_map$fopr (a):<cloref1> b

[Hongwei Xi]

stream_vt_con is considered an internal type; it should be kept as such.

>> fun {a,b:vt@ype} stream_vt_map$fopr (a -<cloref1> b): b

This is probably easier to use but less efficient when [a] is large. In C, one rarely passes a big struct; always

passes a pointer to a big struct.

stream_vt is very different from stream; stream_vt is essentially just one node (while stream is a list of nodes).

So when using stream_vt, one may use stream_vt(T) for a very big flat T.

On Thu, Jul 14, 2016 at 1:40 PM, Steinway Wu <stein...@gmail.com> wrote:

What about this? Let the worker function consume a. And in map, free the constructor, instead of unfold the constructor. 

fun {a,b:vt@ype} stream_vt_map_con (xs: stream_vt a):

stream_vt_con b
fun {a,b:vt@ype} stream_vt_map$fopr (a -<cloref1> b): b

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[gmhwxi]

Maybe what your want is some kind of mapfree function?

Please take a look at list_vt_mapfree.

For stream_vt, it makes sense to process a linear stream until some condition is met
and then process the rest of the stream later.

[Yannick Duchêne]

Le jeudi 14 juillet 2016 16:39:08 UTC+2, gmhwxi a écrit :

1.

When [a] is a linear type, a?! is only the data part of [a] (that is, there is no view in a?!).

When [a] is non-linear, then a?! is just [a].

I thought `a?!` was meaning “initialized” . The data part: does it means `a?!` is of some datatype ?

[Hongwei Xi]

Conceptually, a view VT equals (V | T), where V is the view and T is the data.

So VT?! equals T.

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[Yannick Duchêne]

So if `?!` means “data part” may be it’s better to read `?` as “no data part” instead of “unintitialized”, although it may be implied.

I just wonder where the wording topization (for `?`) and typization (for `?!`) come from. That ’s the wording used in Postiats source. I though about the logical “top” symbol, but it’s unrelated I believe.

[Hongwei Xi]

Given a type T, T is always a subtype of T?. That is why the word 'top' is used.

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[Yannick Duchêne]

Le samedi 23 juin 2018 11:51:53 UTC+2, Yannick Duchêne a écrit :

So if `?!` means “data part” may be it’s better to read `?` as “no data part” instead of “unintitialized”, although it may be implied.

[…]

Or may be both. It seems `_`  as a dynamic expression becomes a `D1Etop` (top, again) in Postiats, and a comment about it somewhere says “uninitialized exp”.

I wanted to check something after I learned this, and was surprised to see this:

        static fn f (int): int

        val a = f(_)

… type‑checks.

[Hongwei Xi]

The symbol '_' refers to a "special" value that is unspecified.

This feature is not yet implemented in ATS2 (and likely will not

in future, either).

If you compile the above code, you will get an error message of

some kind.

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