Dynamic Semantics

COS 441

Princeton University

Fall 2004

Outline for Today

• Evaluation/Big-Step/Natural Semantics– Semantics for -calculus– SML interpreter for -calculus

• Transition/Small-Step/SOS Semantics– Semantics for -calculus– SML interpreter for -calculus

• Harper covers these two topics in the other order

• Relationship between the two approaches

Evaluation Semantics

• Evaluation semantics requires the definition of two sets– evaluatable expressions (E)– values (V)

• Useful but not required that V ½ E

• Defines a binary relation ½ E £ V

Static Semantics for -calculus

Names x 2 …Expressions e ::= lam(x.e) | apply(e1,e2)| x

` X okX 2

ok-V

` apply(E1,E2)ok

` E1 ok ` E2 okok-A

` lam(X.E)ok

[ {X} ` E okok-L

Evaluation Semantics for -calculus

Lambda Expr’s E = { E | {} ` E ok }

Lambda Values V = { lam(X.E) | lam(X.E) 2 E }

apply(E1,E2) V

E1 lam(X’.E’) E2 V’ [X’ Ã V’] E’ V eval-A

lam(X.E)lam(X.E)

lam(X.E) 2 V eval-L

Evaluation Semantics for -calculus

Substitution of a closed term V’ into E’ is closed

Lambda Expr’s E = { E | {} ` E ok }

Lambda Values V = { lam(X.E) | lam(X.E) 2 E }

apply(E1,E2) V

E1 lam(X’.E’) E2 V’ [X’ Ã V’] E’ V eval-A

lam(X.E)lam(X.E)

lam(X.E) 2 V eval-L

Example: Evaluations

apply(lam(X.X),lam(Y.Y)) ??

apply(lam(X.Z),lam(Y.Y)) ??

lam(X.X) ??

apply(X,lam(Y.Y)) ??

Example: Evaluations

apply(lam(X.X),lam(Y.Y)) ??

apply(lam(X.Z),lam(Y.Y)) ??

lam(X.X) lam(X.X)

apply(X,lam(Y.Y)) ??

Example: Evaluations

apply(lam(X.X),lam(Y.Y)) lam(Y.Y)

apply(lam(X.Z),lam(Y.Y)) ??

lam(X.X) lam(X.X)

apply(X,lam(Y.Y)) ??

Example: Evaluations

apply(lam(X.X),lam(Y.Y)) lam(Y.Y)

apply(lam(X.Z),lam(Y.Y)) ??

lam(X.X) lam(X.X)

apply(X,lam(Y.Y)) ??

Example: Evaluations

apply(lam(X.X),lam(Y.Y)) lam(Y.Y)

apply(lam(X.Z),lam(Y.Y)) ??

lam(X.X) lam(X.X)

apply(X,lam(Y.Y)) ??

Example: Evaluations

apply(lam(X.apply(X,X)), lam(X.apply(X,X)) ??

Example: Evaluations

apply(lam(X.apply(X,X)), lam(X.apply(X,X)) ??

[XÃlam(X.apply(X,X))] lam(X.apply(X,X)) apply(lam(X.apply(X,X)),

lam(X.apply(X,X))

Example: Evaluations

There is no derivation because the expression “infinite loops”

apply(lam(X.apply(X,X)), lam(X.apply(X,X)) ??

[XÃlam(X.apply(X,X))] lam(X.apply(X,X)) apply(lam(X.apply(X,X)),

lam(X.apply(X,X))

Evaluation “Order”

• Which subterm do we evaluate first?

• Rules don’t specify, you can build your proof tree up in any order!

• Order of evaluation doesn’t really matter in this case

apply(E1,E2)

Evaluation Semantics in SML

apply(E1,E2) V

E1 lam(X’.E’) E2 V’ [X’ Ã V’] E’ V eval-A

lam(X.E)lam(X.E)

lam(X.E) 2 V eval-L

Evaluation Semantics in SML eval(lam(X.E)) lam(X.E)

if lam(X.E) 2 V

apply(E1,E2) V

E1 lam(X’.E’) E2 V’ [X’ Ã V’] E’ V eval-A

Evaluation Semantics in SML eval(lam(X.E)) lam(X.E)

if lam(X.E) 2 V

eval(apply(E1,E2)) eval([X’ÃV’]E’))

if lam(X’.E’)eval(E1))

and V’ eval(E2)

Evaluation Semantics in SMLtype name = Names.namedatatype exp = Lam of (name * exp) | Apply of (exp * exp) | Var of (name)

val subst : (name * exp * exp) -> exp val ok : (Names.set * exp) -> bool

Evaluation Semantics in SML

fun eval(Lam(x,e)) = if ok(Set.empty,Lam(x,e)) then Lam(x,e) else raise Wrong | eval(Apply(e1,e2)) = let val Lam(x’,e’) = eval(e1) val v’ = eval(e2) in eval(subst(x,v’,e’)) end

Evaluation Semantics in SML

fun eval(Lam(x,e)) = if ok(Set.empty,Lam(x,e)) then Lam(x,e) else raise Wrong | eval(Apply(e1,e2)) = (case(eval(e1),eval(e2)) of (Lam(x’,e’),v’) =>

eval(subst(x,v’,e’)) | _ => raise Impossible) | eval _ = raise Wrong

Evaluation Semantics in SML

fun eval(e as Lam _) = if ok(Set.empty,e)then e else raise Wrong | eval(Apply(e1,e2)) = (case(eval(e1),eval(e2)) of (Lam(x’,e’),v’) =>

eval(subst(x’,v’,e’)) | _ => raise Impossible) | eval _ = raise Wrong

Another Approach

Introduce an type that represents valuestype name = Names.name

datatype exp = Lam of (name * exp) | Apply of (exp * exp) | Var of (name) type value = (name * exp)

Another Approach

fun eval(Lam(x,e)) = if ok(Set.empty,Lam(x,e)) then (x,e) else raise Wrong | eval(Apply(e1,e2)) = let val (x’,e’) = eval(e1) val v’ = eval(e2) in eval(subst(x,Lam v’,e’)) end | eval _ = raise Wrong

Evaluation Semantics

• Non-termination expressions do not evaluate to a value

• Evaluation “order” unspecified

• Concise and “intuitive”

• Easy to translate into efficient interpreter– The main reason we study it!

Transition Semantics

• Transition semantics requires the definition of universe of states S– initial states (I)– final states (F)

• Required that I ½ S and F ½ S

• A binary step relation ½ S £ S

• Iterated multi-step versions *,!, and n

Multistep Relations

S * SZ*

S * S’’

S S’ S’ * S’’S*

S ! S

S 2 FZ!

S ! S’’

S S’ S’ ! S’’S!

S 0 SZn

S n+1 S’’

S S’ S’ n S’’Sn

Transition Semantics for -calculus

((x.e1) e2) ((x.e1) e’2)

e2 e’2A2

((x.e1) (y.e2)) [xÃ(y.e2)] e1 A1

(e1 e2) (e’1 e2)

e1 e’1 A3

I = { E | {} ` E ok }

S = { E | 9. ` E ok }

F = { x.e | {} ` x.e ok }

Example:

((x.x) (y.y)) ??

x.x) ??

((x.(x x)) (x.(x x))) ??

(((x.x) (x.x)) ((x.x) (x.x))) ??

Example:

((x.x) (y.y)) ??

x.x) ??

((x.(x x)) (x.(x x))) ??

(((x.x) (x.x)) ((x.x) (x.x))) ??

Example:

((x.x) (y.y)) (x.x)

x.x) ??

((x.(x x)) (x.(x x))) ??

(((x.x) (x.x)) ((x.x) (x.x))) ??

Example:

((x.x) (y.y)) (x.x)

x.x) ??

((x.(x x)) (x.(x x))) ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) ??

Example:

((x.x) (y.y)) (x.x)

x.x) ??

((x.(x x)) (x.(x x))) ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) ((x.x) ((x.x) (x.x)))

Example:

((x.x) (y.y)) (x.x)

x.x) ??

((x.(x x)) (x.(x x))) ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) ((x.x) ((x.x) (x.x)))

((x.x) ((x.x) (x.x))) ??

Example:

((x.x) (y.y)) (x.x)

x.x) ??

((x.(x x)) (x.(x x))) ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) ((x.x) ((x.x) (x.x)))

((x.x) ((x.x) (x.x))) ((x.x) (x.x))

Example:

((x.x) (y.y)) (x.x)

x.x) ??

((x.(x x)) (x.(x x))) ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) ((x.x) ((x.x) (x.x)))

((x.x) ((x.x) (x.x))) ((x.x) (x.x))

((x.x) (x.x)) ??

Example:

((x.x) (y.y)) (x.x)

x.x) ??

((x.(x x)) (x.(x x))) ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) ((x.x) ((x.x) (x.x)))

((x.x) ((x.x) (x.x))) ((x.x) (x.x))

((x.x) (x.x)) (x.x)

Example: *

((x.x) (y.y)) * ??

x.x) * ??

((x.(x x)) (x.(x x))) * ??

(((x.x) (x.x)) ((x.x) (x.x))) * ??

Example: *

((x.x) (y.y)) * ((x.x) (y.y))

x.x) * x.x)

((x.(x x)) (x.(x x))) * ??

(((x.x) (x.x)) ((x.x) (x.x))) * ??

Example: *

((x.x) (y.y)) * (y.y)

x.x) * x.x)

((x.(x x)) (x.(x x))) * ??

(((x.x) (x.x)) ((x.x) (x.x))) * ??

Example: *

((x.x) (y.y)) * (y.y)

x.x) * x.x)

((x.(x x)) (x.(x x))) * ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) * ??

Example: *

((x.x) (y.y)) * (y.y)

x.x) * x.x)

((x.(x x)) (x.(x x))) * ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) * (((x.x) (x.x)) ((x.x) (x.x)))

Example: *

((x.x) (y.y)) * (y.y)

x.x) * x.x)

((x.(x x)) (x.(x x))) * ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) * ((x.x) ((x.x) (x.x)))

Example: *

((x.x) (y.y)) * (y.y)

x.x) * x.x)

((x.(x x)) (x.(x x))) * ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) * ((x.x) (x.x))

Example: *

((x.x) (y.y)) * (y.y)

x.x) * x.x)

((x.(x x)) (x.(x x))) * ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) * (x.x)

Example: !

((x.x) (y.y)) ! ??

x.x) ! ??

((x.(x x)) (x.(x x))) ! ??

(((x.x) (x.x)) ((x.x) (x.x))) ! ??

Example: !

((x.x) (y.y)) ! ??

x.x) ! x.x)

((x.(x x)) (x.(x x))) ! ??

(((x.x) (x.x)) ((x.x) (x.x))) ! ??

Example: !

((x.x) (y.y)) ! (y.y)

x.x) ! x.x)

((x.(x x)) (x.(x x))) ! ??

(((x.x) (x.x)) ((x.x) (x.x))) ! ??

Example: !

((x.x) (y.y)) ! (y.y)

x.x) ! x.x)

((x.(x x)) (x.(x x))) ! ??

(((x.x) (x.x)) ((x.x) (x.x))) ! ??

Example: !

((x.x) (y.y)) ! (y.y)

x.x) ! x.x)

((x.(x x)) (x.(x x))) ! ??

(((x.x) (x.x)) ((x.x) (x.x))) ! (x.x)

Example: 0

((x.x) (y.y)) 0 ??

x.x) 0 ??

((x.(x x)) (x.(x x))) 0 ??

(((x.x) (x.x)) ((x.x) (x.x))) 0 ??

Example: 0

((x.x) (y.y)) 0 ((x.x) (y.y))

x.x) 0 x.x)

((x.(x x)) (x.(x x))) 0 ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) 0 (((x.x) (x.x)) ((x.x) (x.x)))

Example: 1

((x.x) (y.y)) 1 (y.y)

x.x) 1 ??

((x.(x x)) (x.(x x))) 1 ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) 1 ((x.x) ((x.x) (x.x)))

Example: 2

((x.x) (y.y)) 2 ??

x.x) 2 ??

((x.(x x)) (x.(x x))) 2 ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) 2 ((x.x) (x.x))

Example: 3

((x.x) (y.y)) 3 ??

x.x) 3 ??

((x.(x x)) (x.(x x))) 3 ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) 3 (x.x)

Example: 4

((x.x) (y.y)) 4 ??

x.x) 4 ??

((x.(x x)) (x.(x x))) 4 ((x.(x x)) (x.(x x)))

(((x.x) (x.x)) ((x.x) (x.x))) 4 (x.x)

!

((x.x) (y.y)) ! (y.y)

x.x) ! x.x)

((x.(x x)) (x.(x x))) ! ??

(((x.x) (x.x)) ((x.x) (x.x))) ! (x.x)

((x.x) (y.y)) (y.y)

x.x) x.x)

((x.(x x)) (x.(x x))) ??

(((x.x) (x.x)) ((x.x) (x.x))) (x.x)

Relation Between ! and

Theorem: e ! (x.e’) iff e (x.e’)Proof: By lemma 1 and lemma 2

Lemma 1: If e (x.e’) then e ! (x.e’)

Lemma 2: If e ! (x.e’) then e (x.e’)Proofs: See Harper 7.3 for sketch

Transition Semantics

• Has direct correspondence with evaluation semantics

• Let us reason about– Evaluation order– Computational complexity of program– Partial computations

• Easier to relate with static semantics to prove type safety– Next lecture!