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.
On Thu, Jul 14, 2016 at 9:20 AM, Steinway Wu <steinway...@gmail.com> wrote:
> 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,
>
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