midterm & concept review

Midterm & Concept Review

CS 510, Fall 2002David Walker

Midterm

Most people did just fine greater than 90 = exceptional 80-90 = good less than 80 = need more work

Three main parts: MaybeML: 35 Objects: 35 Imperative objects: 30

Type-directed TranslationsWhat is a type-directed translation?

Type-directed TranslationsWhat is a type-directed translation? it is a function (a compiler) that takes a

typing derivation for a source-language expression and produces a target-language expression

we can define type-directed translations using judgments:

(G |- e : t) ==> e’ e is a source term; e’ is a target term we define the translation using inference rules. eg:(G |- e1 : t1 -> t2) ==> e1’ (G |- e2 : t1) ==> e2’

-------------------------------------------------------------------(G |- e1 e2 : t2) ==> e1’ e2’

Type-preserving Translations

When is a translation is type-preserving?

Type-preserving Translations

When is a translation is type-preserving? If given a valid derivation, it produces a

well-typed target expression We often prove a theorem like this:

if (G |- e : t) ==> e’ then Trans(G) |- e’ : Trans(t)

where Trans(t) is atype translation function

this is an ordinary typingjudgmentin the targetlanguage

Type Translation

What I expected: Trans (Bool?) = unit + (unit + unit) Trans (t1 ?-> t2) = unit + (Trans(t1) + Trans(t2))

Another possibility: Trans (Bool?) = unit + (unit + unit) Trans (t1 ?-> t2) = t1 -> t2

Type Translation

Almost: Trans (Bool?) = anyt Trans (t1 ?-> t2) = anyt

where anyt = unit + (unit + unit) + (anyt -> anyt)

what is wrong here?

Type Translation

Almost: Trans (Bool?) = anyt Trans (t1 ?-> t2) = anyt

where anyt = unit + (unit + unit) + (anyt -> anyt)

what is wrong here? Need a recursive type: anyt = rec a. unit + (unit + unit) + (a -> a)

Term Translation

Mirrors the form of the static semanticsUses judgments with the form:

(G |- e1 : t) ==> e1’

Invariant: If (G |- e1 : t) ==> e1’ then

Trans(G) |- e1’ : Trans(t)Key: e1’ must type check underfully translated environment

Term Translation (no opt.)

---------------------------------------------------------------------------(G |- true : Bool?) ==> inr (Trans(Bool?), inl (unit + unit, ()))

(G |- e1 : t1 ?-> t2) ==> e1’(G |- e2 : t1) ==> e2’ (x,y not in Dom(G))---------------------------------------------------------------------------(G |- e1 e2 : t2) ==> case e1’ of ( inl (x) => fail | inr (y) => y e2’)

Example:

Example:

Wrong (but it was pretty tricky, so don’t worry):


(G,f : t1 ?-> t2, x : t1 |- e : t2) ==> e’ ------------------------------------------------(G |- fun f (x : t1) : t2 = e ==> inr (..., fun f (x : Trans(t1)) : Trans(t2) = e’)

What goes wrong?

Consider:fun f (x : unit) : unit = f x

and its translation:inr (..., fun f (x : unit) : unit = case f of ( inl (y) => fail | inr (z) => z x))

thisis a functionNOT a sum value

The point of doing a proof is to discover mistakes!

Must prove result of trans has the right type:


------------------------------------------------------ (by IH)G,f : T(t1) -> T(t2), x : T(t1) |- e’ : T(t2) ------------------------------------------------------------------ (fun)Trans(G) |- fun f (x : T(t1)) : T(t2) = e’ : T(t1) -> T(t2)------------------------------------------------------------------ (inr)Trans(G) |- inr (..., ....) : unit + (T(t1) -> T(t2))

We can’t apply the induction hypothesis! (T(G) |- e : t) ==> e’ is necessary


------------------------------------------------------ (by IH)G,f : T(t1) -> T(t2), x : T(t1) |- e’ : T(t2) ------------------------------------------------------------------ (fun)Trans(G) |- fun f (x : T(t1)) : T(t2) = e’ : T(t1) -> T(t2)------------------------------------------------------------------ (inr)Trans(G) |- inr (..., ....) : unit + (T(t1) -> T(t2))

not the translation of a function type (unit + T(t1) -> T(t2))

How do we fix this? create something with type (unit + T(t1) ->

T(t2)) to use inside the function then bind that something to a variable f for use

inside e’ our old translation gave us something with the

type T(t1) -> T(t2) ...


How do we fix this?


(G,f : t1 ?-> t2, x : t1 |- e : t2) ==> e’ ------------------------------------------------(G |- fun f (x : t1) : t2 = e ==> inr (..., fun f (x : Trans(t1)) : Trans(t2) = let f = inr (..., f) in e’)

wherelet x = e1 in e2 is ((fun _ (x : ...) : ... = e2) e1)

A useful fact

A let expression is normally just “syntactic sugar” for a function application let x = e1 in e2is the same as (fn x => e2) e1

Optimization

Observation: There are only a couple of null checks

that appear in our translation. Can we really do substantially better?

Optimization

Observation: There are only a couple of null checks

that appear in our translation. Can we really do substantially better?

YES! The expressions that result from the

translation have many, many, many null checks:

if true then if false then if true then ....

our translation inserts 3 unnecessary null checks!

Some Possibilities

Some people defined a few special-case rules: Detect cases where values are directly

elimiated and avoid null checks: (fun f (x:t1) :t2 = e) e’ translated differently if true then e1 else e2 translated differently

Others: if? x then (if? x then e1 else e2) else e3

These people received some marks

A Much More General Solution

Do lazy injections into the sum type Keep track of whether or not you have done

the injection using the type of the result expression

if t’ = (unit + unit) or (trans(t1) -> trans(t2)) then you haven’t injected the expression e’ into a sum yet

you can leave off unnecessary injections around any expression, not just values

(G |- e : t) ==> (e’ : t’)

A Much More General Solution

New rules for introduction forms:

Extra rules for elimination forms:

----------------------------------------------------------(G |- true : Bool?) ==> (inr (..., ()): unit + unit)

(G |- e1 : Bool? ==> (e1’, t) t notnull(G |- e2 : ts ==> (e2’, t2’)(G |- e3 : ts ==> (e3’, t3’) e2’,e3’,t2’,t3’ unifies to e2’’,e3’’,t’’----------------------------------------------------------(G |- if e1 then e2 else e3 : ts) ==> if e1 then e2’’ else e3’’ : t’’

New Judgments

Natural Types:

“Unification”

------------------------(unit + unit) notnull

----------------------(t1 -> t2) notnull

t2 = t3-------------------------------------e2,e3,t2,t3 unifies to e2,e3,t2

t2 = unit + t3---------------------------------------------e2,e3,t2,t3 unifies to e2,inr(t2,e3),t2

t3 = unit + t2---------------------------------------------e2,e3,t2,t3 unifies to inr(t3,e2),e3,t3

Objects

Most people did well on the definition of the static and dynamic semantics for objects if you want to know some detail,

come see me during my office hours

Less well on the imperative features

Terminology What is a closed expression?

Terminology What is a closed expression?

An expression containing no free variables.

((fun f (x:bool):bool = x x) true) is closed ((fun f (x:bool):bool = y x) true) is not

closed Mathematically: if FV is a function that

computes the set of free variables of an expression then

e is closed if and only if FV(e) = { }

Terminology What is a well-formed expression?

Terminology What is a well-formed expression?

An expression that type checks under some type context G.

((fun f (x:bool):bool = x x) true) is not well formed

((fun f (x:bool):bool = y x) true) is well-formed in the context G = [y:bool -> bool]

Terminology

What is (e : t) an abbreviation for?

Terminology

What is (e : t) an abbreviation for? the typing judgment:

. |- e : t

If (e : t) then what else do we know?

empty context

Terminology

What is (e : t) an abbreviation for? the typing judgment:

. |- e : t

If (e : t) then what else do we know? we know that e contains no free variables in other words, e is closed (we might know other things if e also

happens to be a value)

empty context

Terminology What is a value?

Terminology What is a value?

it is an expression that does not need to be further evaluated (and it is not stuck)

How do we normally define values?

Terminology How do we normally define values?

We declare a new metavariable v and give its form using BNF:

v ::= x | n | <v1,v2> | fun f (x : t1) : t2 = e What is the difference between a

metavariable v and an expression variable x?

Alternatively, we define a value judgment:

--------------|- n value

--------------|- x value

|- v1 value |- v2 value----------------------------------|- <v1,v2> value

Terminology

Does it matter whether we use BNF or a series of judgments to define the syntax of values and expressions?

Terminology

Does it matter whether we use BNF or a series of judgments to define the syntax of values and expressions? No! BNF is just an abbreviation for the

inductive definition that we would give using judgments instead

Why don’t we define typing rules using BNF if it is so darn convenient?

Terminology

Does it matter whether we use BNF or a series of judgments to define the syntax of values and expressions? No! BNF is just an abbreviation for the

inductive definition that we would give using judgments instead

Why don’t we define typing rules using BNF if it is so darn convenient? Typing rules are context-sensitive. BNF is

used for context-insensitive definitions.

Terminology What is strange about the following

sentence? If (v : t) and v is a closed, well-formed

value then the canonical forms lemma can tell us something about the shape of v given the type t.

Terminology What is strange about the following

sentence? If (v : t) and v is a closed, well-formed value then

the canonical forms lemma can tell us something about the shape of v given the type t.

The red part is totally redundant! If you are using the metavariable v, then you should

have already defined it so that it refers to values. (v : t) should also have been defined before. It should

trivially imply that v is closed. It defines what it means for v to be well-formed!

If you write a sentence like this on the final, you might find yourself losing points....

Back to objects

In the future, when I say “write an expression that does ...” you should always write a well-formed, closed expression unless I specify otherwise.

{getloop =

fn (x). ({loop = fn(y).y.loop} : {loop : t})

} : {getloop : {loop : t} }

isn’t really an expression! It contains the metavariable t.

Back to objects

{getloop = fn (x). ({loop = fn(y).y.loop} : {loop : { }}) } : {getloop : {loop : { }} }

is what you want to do.

Imperative objects

Syntax t ::= {l = t,...} e ::= x | {l = b,...} | e.l | e.l <- b | ... b ::= fn(x).e

Operational semantics

Without imperative features (field update) we can use the ordinary M-machine definitions

e -> e’


The obvious M-machine definition for object update doesn’t work:

e = {lk = b’,l’’ = b’’...}-----------------------------------------(e.lk <- b) -> {lk = b,l’’ = b’’...}


let x = {n = fn(_).3} inlet _ = (x.n <- fn(_).2) inx.n

let _ = ({n = fn(_).3}.n <- fn(_).2) in{n = fn(_).3}.n

let _ = {n = fn(_).2} in{n = fn(_).3}.n

{n = fn(_).3}.n 3

Here’s why: this updateonly has local effect

Operational semanticsWe need to augment our operational semantics with a global store.A store S is a finite partial map from locations (r) to values. (what is a finite partial map?) v ::= {l=b,...} | r run-time expressions include locations r

Our semantics now has form: (S,e) -> (S’,e’)


Rules:

------------------------------------------------(S, {l = b,...}) -> (S[r -> {l = b,...}], r)

S(r) = {l = fn(x).e,...}-----------------------------(S, r.l) -> (S, e[S(r)/x])

S(r) = {l = b’’, l’ = b’,...}-------------------------------------------------------(S, r.l <- b) -> (S[r -> {l = b, l’ = b’,...}], r)


(., let x = {n = fn(_).3} in let _ = (x.n <- fn(_).2) in x.n)

([r -> {n = fn(_).3}], let _ = (r.n <- fn(_).2) in r.n)

Our example:

([r -> {n = fn(_).2}], r.n)

([r -> {n = fn(_).2}], 2)

emptystore

r is substitutedfor x everywherebut the contentsof r are kept in one place

Summary

Things to remember: how to define type-directed and type-

preserving translations be able to use and define common terms

values, closed expressions, operational semantics, canonical form, inversion principle, type system, soundness, completeness, subtyping, the subsumption principle, etc.

proofs are for finding mistakes imperative features are tricky

midterm & concept review

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