In the provided example, why is it that we can replace zero with any
natural number?
...
free@{a}{123456(*insert random nat here*)}(xs);...
...
(typechecks and compiles with no issues...)
On Sunday, February 10, 2019 at 8:43:22 PM UTC-5, gmhwxi wrote:
>
>
> A list node contains two cells: one holding the content and the other
> holding
> the pointer to the next node (or null).
>
> When xs matches the pattern @list_vt_cons(x, xs1), xs refers to a list
> node;
> x refers to the content cell and xs1 refers to the pointer cell. In this
> case, the
> content is non-linear (a:t@ype) and thus does not need to be moved out. But
> the pointer needs to be moved out before xs can be freed:
>
> val xs1_ = xs
>
> The above code was written long time ago. It could be prettified a bit as
> follows:
>
> fun
> {a:t@ype}
> list_vt_free
> {n:nat} .<n>.
> (xs: list_vt(a, n)): void =
> (
> case+ xs of
> | ~list_vt_nil() => ()
> | @list_vt_cons(_, xs1) =>
> let
> val xs1 = xs1
> in
> free@{a}{0}(xs); list_vt_free<a>(xs1)
> end
> )
>
> People are often puzzled by a beautiful line like 'val xs1 = xs1' :)
>
> On Sunday, February 10, 2019 at 6:17:03 PM UTC-5, rnagasam wrote:
>>
>> Hi all,
>>
>> I'm facing some difficulty understanding how `free@' is used. I'm going
>> through the *Introduction to Programming in ATS* book, and the example
>> I'm facing trouble with is this,
>>
>> fun{
>> a:t@ype
>> } list_vt_free
>> {n:nat}.<n>. (xs: list_vt (a, n)): void =
>> case+ xs of
>> | @list_vt_cons(x, xs1) => let
>> val xs1_ = xs1
>> val () = free@{a}{0}(xs) in list_vt_free (xs1_)
>> end
>> | @list_vt_nil () => free@{a} (xs)
>>
>> Why is it `free@{a}{0} (xs)' and not `free@{a}{0} (x)'? It seems that
>> `free@' requires an unfolded constructor (here, it is operating on
>> `list_vt_cons_unfold'). However, that begs the question -- how does `free@'
>> know to free only the first node of the list and not the rest of the list?
>> Any help understanding this will be much appreciated!
>>
>> Thanks,
>> Ramana
>>
>
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