(set-box! ycell 3)
(+ x (unbox ycell)))
-(C has explicit-style mutable variables, too, which it calls *pointers*. But simple variables in C are already mutable, in the implicit style.)
+C has explicit-style mutable variables, too, which it calls *pointers*. But simple variables in C are already mutable, in the implicit style. Scheme also has both styles of mutation. In addition to the explicit boxes, Scheme also lets you mutate unboxed variables:
+
+ (begin
+ (define x 1)
+ (set! x 2)
+ x)
+ ; evaluates to 2
When dealing with explicit-style mutation, there's a difference between the types and values of `ycell` and `!ycell` (or in Scheme, `(unbox ycell)`). The former has the type `int ref`: the variable `ycell` is assigned a reference cell that contains an `int`. The latter has the type `int`, and has whatever value is now stored in the relevant reference cell. In an implicit-style framework though, we only have the resources to refer to the contents of the relevant reference cell. `y` in fragment [G] or the C snippet above has the type `int`, and only ever evaluates to `int` values.
Let's consider how to interpet our new syntactic forms `newref`, `deref`, and `setref`:
-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:
+1. When `expr` evaluates to starting\_val, **newref expr** 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 ...
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 (assuming it wasn't changed during the evaluation of `expr`).
+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 (assuming it wasn't changed during the evaluation of `expr`).
let rec eval expression g s =
match expression with
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`.
+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