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c3527b5)
Signed-off-by: Jim Pryor <profjim@jimpryor.net>
Let's consider how to interpet our new syntactic forms `newref`, `deref`, and `setref`:
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. **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 ...
let ycell = newref 1
in ...
in (Index new_index, s'')
...
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
let rec eval expression g s =
match expression with
in (List.nth s' n, s')
...
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
let rec eval expression g s =
match expression with