CS 334
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We normally expect that if we change the names of formal parameters that it should not make any difference, but ...
Suppose we evaluate:
let fun g x y = x + y in g y end;(or in our language PCF:
(fn g => g y) (fn x => fn y => x+y))If we evaluate blindly we get:
fn y => y + yNotice that because of scoping, the actual parameter y has become captured by the formal parameter y!
We should get: fn w => y + w, which has a very different meaning!!
(Note that we did not run into this problem earlier since during our evaluations we never worked with terms with free variables - when going inside functions we replaced all formal parameters by the actual parameters, which didn't involve free variables).
A different order of evaluation would have brought forth the same problem, however.
We would like to have fn x => B to represent the same function as fn y => B[x:=y] as long as y doesn't occur freely in B. (Called alpha-conversion)
If you always alpha-convert to a new bound variable before substituting in, will never have problems, but this is a pain in the neck.
Instead we will valuate terms with respect to environments. Intuitively, an environment is a mapping from strings (representing identifiers) to values. That is, it tells us for every identifier currently in scope, what the value of that identifier is.
Rather than representing the environment as a function, we will represent it as an "association list", a set of pairs of strings and values:
env = (string * value) listWe can look up the value of an identifier in an environment by simply searching for the first occurrence of the identifier in the list.
We write [[e]] ev for the meaning of e with respect to environment ev.
E.g. if ev(x) = 12 and ev(y) = 2, then [[x+y]] ev = 14.
How does function application result in change of environment?
[[(fn x => body) actual]]ev = [[body]] (ev [x := [[actual]]ev])where ev[x := v] is environment like ev except x has value "v".
This and rec are the only rules in which the environment changes!
Rest of rules look like the old interpreter (except identifiers are now looked up in the environment)!
Replaces all uses of subst!
This means that computation no longer takes place by rewriting terms into new terms, interp is now a function from term to value.
Note that
let val x = arg in eis equivalent to
(fn x => e) argMust worry about scoping problems:
val test = let val x = 3; fun f y = x + y; val x = 12 in x + (f 7) end;What is value of test? In particular, what is the value of (f 7)?
Change in scope is reflected by change in environment.
With functions must remember environment function was defined in!
When apply function, apply in defining environment.
test is equivalent to
(fn x => (fn f => ((fn x => x + (f 7)) 12) (fn y => x + y))) 3Then
[[(fn x => (fn f => ((fn x => x + (f 7)) 12) (fn y => x + y))) 3]] ev0 = [[(fn f => ((fn x => x + (f 7)) 12) (fn y => x + y)) ]] ev1 = [[(fn x => x + (f 7)) 12]] ev2 = [[x + (f 7)]] ev3 = 12 + ([[fn y => 3 + y]] ev1) 7 = 12 + [[3 + y]] ev4 = 12 + 3 + 7 = 22where ev0 is the starting environment and
ev1 = ev0 [x := [[3]] ev0] = ev0[x := 3] ev2 = ev1 [f := [[fn y => x + y]] ev1] <- Closure for f ev3 = ev2 [x := [[12]] ev2] = ev2[x := 12] ev4 = ev1 [y := 7]Notice that ev2 is created by adding a closure to the environment to represent the meaning of f. In the formal language, we would write that particular closure more like CLOSURE ("y", x + y, ev1). (The only thing we can't do directly in the language is define x + y.
Mapping from index type to range type
E.g. Array [1..10] of Real corresponds to {1,...,10} ->
Real
Operations and relations: selection ". [.]", =, ==, and occasionally slices.
E.g. A[2..6] represents an array composed of A[2] to A[6]
Index range and location where array stored can be bound at compile time, unit activation, or any time.
The key to these differences is binding time, as usual!
What is difference from an array? Efficiency, esp. w/update.
update f arg result x = if x = arg then result else f xor
update f arg result = fn x => if x = arg then result else f xProcedure can be treated as having type S -> unit for uniformity.
tree = Empty | Mktree of int * tree * treeIn most lang's built by programmer from pointer types.list = Nil | Cons of int * list
Sometimes supported by language (e.g. Miranda, Haskell, ML).
Why can't we have direct recursive types in ordinary imperative languages?
OK if use ref's:
list = POINTER TO RECORD first:integer; rest: list END;
Recursive types may have many sol'ns
E.g. list = {Nil} union (int x list) has following sol'ns:
Theoretical result: Recursive equations always have a least solution - though infinite set if real recursion.
Can get via finite approximation. I.e.,
list0 = {Nil}Very much like unwinding definition of recursive functionlist1 = {Nil} union (int x list0) = {Nil} union {(n, Nil) | n in int}
list2 = {Nil} union (int x list1) = {Nil} union {(n, Nil) | n in int} union {(m,(n, Nil)) | m, n in int}
...
list = Unionn listn
fact = fun n => if n = 0 then 1 else n * fact (n-1) fact0 = fun n => if n = 0 then 1 else undef fact1 = fun n => if n = 0 then 1 else n * fact0(n-1) = fun n => if n = 0, 1 then 1 else undef fact2 = fun n => if n = 0 then 1 else n * fact1(n-1) = fun n => if n = 0, 1 then 1 else if n = 2 then 2 else undef ... fact = Unionn factnNotice solution to T = A + (T->T) is inconsistent with classical mathematics!
datatype univ = Base of int | Func of (univ -> univ);
operations: hd, tail, cons, length, etc.
Persistent data - files.
Are strings primitive or composite?
var x : integer {bound at translation time}The variable can only hold values of that type. (Pascal/Modula-2/C, etc.)
FORTRAN has implicit declaration using naming conventions
In either case, run real danger of problems due to typos.
Example in ML, if
datatype Stack ::= Nil | Push of int;then define
fun f Push 7 = ...What error occurs?
Answer: Push is taken as a parameter name, not a constructor.
Therefore f is given type: A -> int -> B rather than the
expected: Stack -> B
Dynamic: Variables typically do not have a declared type. Type of value may vary during run-time. Esp. useful w/ heterogeneous lists, etc. (LISP/SCHEME).
Dynamic more flexible, but more overhead since must check type before performing operations (therefore must store tag w/ value).
Dynamic typing found in APL and LISP.
Assignment compatibility:
var x : hex; y : ounces;
Is x := y legal?
Original report said both sides must have identical types.
When are types identical?
Ex.:
Type T = Array [1..10] of Integer; Var A, B : Array [1..10] of Integer; C : Array [1..10] of Integer; D : T; E : T;Which variables have the same type?
--> A, B and D, E only.
Structural not always easy. Let
T1 = record a : integer; b : real end; T2 = record c : integer; d : real end; T3 = record b : real; a : integer end;Which are the same?
Worse:
T = record info : integer; next : ^T end; U = record info : integer; next : ^V end; V = record info : integer; next : ^U end;
Two types are assignment compatible iff they
Things are more complicated in object-oriented languages. Then assignment is OK if type of source is a subtype of the receiver type.
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kim@cs.williams.edu