Re: Questions on closure environment capture

[Quentin Carbonneaux <quentin@xxxxxx>]

On Thu, May 03, 2018 at 12:14:31PM +0800, vimacs wrote:
> Hi,
> 
> I'm trying to use the closure in Myrddin, but I'm confused about the
> implementation of it and the so-called environment capture.
> 
> Here are some code in my program. 

Hi,

Thank your for your very interesting questions.

> So what is the semantic and implementation of closure and environment
> capture? Is there an easy way to implement the functions I said above?

In a nutshell, if you want to do anything "fancy" (i.e., anything that is
not simple Pascal-style nested functions) use std.fndup().

Your example can be made to work by changing it like so:

const f_ = {t
	-> std.fndup({x; -> x+t})
}
[...]
		f[i] = std.fndup({x
			-> x+i
		})

What fndup does is that it takes your closure whose environment
is currently on the *stack* and moves it to the heap.

Since the environements are by default on the stack, your f_()
function cannot work properly because you return a closure whose
environment has been destroyed (because f_ returned).

Now, for the f[i] = {x; -> x+i}.  The reason why you only "see"
the last closure (with i = 3) is that the compiler turns your
code into something like this:

const main = {
	var __myclosure_env
	var i, j
	var f: (x:int -> int)[8]
	var g

	for ...
		__my_closure_env = i
		f[i] = (&__my_closure_env, __code_ptr)
	;;
	[...]
}

As you can see, there is only one local environment for the *syntactic*
closure, not one per dynamically-created closure.  It might be possible
to create one environemnt per *dynamic* closure, but that would require
dynamic stack allocation.  I am not sure what Ori thinks about it.

Let me know if this makes sense.

And Ori, please feel free to correct/enrich the explanation.

Best,

- Q