##How to implement explicit-style mutable variables##
--- FIXME --
+We'll think about how to implement explicit-style mutation first. We suppose that we add some new syntactic forms to a language, let's call them `newref`, `deref`, and `setref`. And now we want to expand the semantics for the language so as to interpret these new forms.
+
+Well, part of our semantic machinery will be an assignment function, call it `g`. Somehow we'd have to keep track of the types of the variables and values we're working with, but we won't pay much attention to that now. In fact, we won't even both much at this point with the assignment function. Below we'll pay more attention to it.
+
+In addition to the assignment function, we'll also need a way to keep track of how many reference cells have been "allocated" (using `newref`), and what their current values are. We'll suppose all the reference cells are organized in a single data structure we'll call a **store**. This might be a big heap of memory. For our purposes, we'll suppose that reference cells only ever contain `int`s, and we'll let the store be a list of `int`s.
+
+In many languages, including OCaml, the first position in a list is indexed `0`, the second is indexed `1` and so on. If a list has length 2, then there won't be any value at index `2`; that will be the "next free location" in the list.
+
+Before we brought mutation on the scene, our language's semantics will have looked something like this:
+
+> \[[expression]]<sub>g</sub> = value
+
+Now we're going to relativize our interpretations not only to the assignment function `g`, but also to the current store, which I'll label `s`. Additionally, we're going to want to allow that evaluating some functions might *change* the store, perhaps by allocating new reference cells or perhaps by updating the contents of some existing cells. So the interpretation of an expression won't just return a value; it will also return a possibly updated store. We'll suppose that our interpretation function does this quite generally, even though for many expressions in the language, the store that's returned will be the same one that the interpretation function started with:
+
+> \[[expression]]<sub>g s</sub> = (value, s')
+
+With that kind of framework, we can interpret `newref`, `deref`, and `setref` as follows.
+
+1. \[[newref starting_val]] should allocate a new reference cell in the store and insert `starting_val` into that cell. It should return some "key" or "index" or "pointer" to the newly created reference cell, so that we can do things like:
+
+ let ycell = newref 1
+ in ...
+
+ and be able to refer back to that cell later by using the value that we assigned to the variable `ycell`. In our simple implementation, we're letting the store just be an `int list`, and we can let the "keys" be indexes in that list, which are just `int`s. Somehow we'd have to keep track of which variables are assigned `int`s as `int`s and which are assigned `int`s as indexes into the store. So we'll create a special type to wrap the latter:
+
+ type store_index = Index of int;;
+
+ let rec eval expression g s =
+ match expression with
+ ...
+ | Newref expr ->
+ let (starting_val, s') = eval expr g s
+ (* note that s' may be different from s, if expr itself contained any mutation operations *)
+ (* now we want to retrieve the next free index in s' *)
+ in let new_index = List.length s'
+ (* now we want to insert starting_val there; the following is an easy but inefficient way to do it *)
+ in let s'' = List.append s' [starting_val]
+ (* now we return a pair of a wrapped new_index, and the new store *)
+ in (Index new_index, s'')
+ ...
+
+2. When `expr` evaluates to a `store_index`, then `deref expr` should evaluate to whatever value is at that index in the current store. (If `expr` evaluates to a value of another type, `deref expr` is undefined.) In this operation, we don't change the store at all; we're just reading from it. So we'll return the same store back unchanged.
+
+ let rec eval expression g s =
+ match expression with
+ ...
+ | Deref expr ->
+ let (Index n, s') = eval expr g s
+ (* note that s' may be different from s, if expr itself contained any mutation operations *)
+ in (List.nth s' n, s')
+ ...
+
+3. When `expr1` evaluates to a `store_index` and `expr2` evaluates to an `int`, then `setref expr1 expr2` should have the effect of changing the store so that the reference cell at that index now contains that `int`. We have to make a decision about what value the `setref ...` call should itself evaluate to; OCaml makes this `()` but other choices are also possible. Here I'll just suppose we've got some appropriate value in the variable `dummy`.
+
+ let rec eval expression g s =
+ match expression with
+ ...
+ | Setref expr1 expr2
+ let (Index n, s') = eval expr1 g s
+ (* note that s' may be different from s, if expr itself contained any mutation operations *)
+ in let (new_value, s'') = eval expr2 g s'
+ (* now we create a list which is just like s'' except it has new_value in index n *)
+ in let rec replace_nth lst m =
+ match lst with
+ | [] -> failwith "list too short"
+ | x::xs when m = 0 -> new_value :: xs
+ | x::xs -> x :: replace_nth xs (m - 1)
+ in let s''' = replace_nth s'' n
+ in (dummy, s''')
+ ...
+
##How to implement implicit-style mutable variables##
in let () = fact_cell := Some factorial
in ...
- We use the `None`/`Some factorial` option type here just as a way to ensure that the contents of fact_cell are of the same type both at the start and the end of the block.
+ We use the `None`/`Some factorial` option type here just as a way to ensure that the contents of `fact_cell` are of the same type both at the start and the end of the block.
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