If you do:

val foo = $extval(struct_foo, "foo")

the foo is a value (a flat record); it cannot be used as a left-value.


On Fri, Aug 19, 2016 at 10:52 PM, Kiwamu Okabe <[email protected]> wrote:

> Hi Hongwei,
>
> On Sat, Aug 20, 2016 at 11:39 AM, gmhwxi <[email protected]> wrote:
> > Please try the following code:
>
> Thanks a lot!
> Totally, `$extval` can only maintain viewtype as pointer. Is it right?
> I'm some confusing...
>
> >  val () = println! !(!pfoo.p) // print 10
>
> I like following style, better. The code is well-typed. My thinking is
> a bad thing?
>
> ```
>   val () = println! !(pfoo->p) // print 10
> ```
>
> Best regards,
> --
> Kiwamu Okabe at METASEPI DESIGN
>
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