Winter 2004/5Pls – expresssion-lang – Catriel Beeri22 Informal Analysis Goal: prove the semantics is deterministic Stages: Values --- we know they have

Winter 2004/5 Pls – expresssion-lang – Catriel Beeri 1

Informal Analysis

Goal: prove the semantics is deterministic

Stages:• Values --- we know they have no transitions• Minimal non-values --- consider the axioms• General composite expressions --- consider the

propagation rules

d


“Minimal” non-value expressions

Reducible expression (redex): • provided it has a transition (condition?)

• (always has a transition) by (op-apply): 3+4 7 by (if_) : if true 3 4 3 Both are reductions

Claim: there is a a unique transition from a redex. (both )

Q: Why “minimal”?

2 3f i b e eo v

, d


Intuition: The rules of define paths in a composite expression E; each descends to a “minimal” non-value sub-expression F

A path defined by the rules: a selection path

The sub-expression at its end: a selected sub-expression

If F is a redex, has a transition F F’, E has a transition E {F’/F}E (replacement)

Arbitrary composite expressions


Examples

(3 5)*(4 5) applic

* tuple

applic applic

+ tuple

3 5

4 5

tuple

(eval-rand)

(eval-tuple) (2,1)

(op-apply)

8*(4 5)


(3 5)*(4 5) applic

* tuple

applic applic

+ tuple

3 5

4 5

tuple

(eval-rand)

(eval-tuple) (2,2)

(op-apply)

(3 5)*( 1)


For , an expression may have several selected sub-expressions, several transitions

But, for d, there is a unique selected sub-expression, hence at most one transition

How do argue this is the case?


Consider application E1 E2• E1 is not a value:

– (op-apply), (eval-rand) cannot be used.

– (eval-rator) can be used,

if E1 has a transition

• E1 is a value, E2 is not:– (op-apply), (eval-rator) cannot be used (why?)

– (eval-rand) can be used,

if E2 has a transition

• Both E1 E2 are values: – (eval-rator),(eval-rand) cannot be used,

– Possibly (op-apply) can be used


This “shows” that for d :For each composite expression E, there is at most one transition

The binary relation d is functional (a partial function)

This transition semantics is deterministic


Natural (evaluation) Semantics

Transition semantics deals with one step

Natural semantics deals with a complete computation

E V

transition

Natural: E V


Natural semantics rules

(nat-int) n n (nat-bool) b b (nat-op) o o

(nat-applic-op) 1 2

1 2 ( )o

e o e v

e e f v

(nat-tuple)1 1

1

...

( ,..., ) ( ,..., )n n

n n n

e v e v

e e v v

(nat-if-true)1 2

1 2 3

true e

if

e v

e e e v

1 3

1 2 3

false e

if

e v

e e e v

(nat-if-false)

[ ]S EL

if defined


A proof tree in natural semantics

Q: What is the difference in shape of trees between transition and natural?

(if 6<2 + )(3,4) 1

6 2 false(if 6<2 + )

3 3 4 4

(3,4) (3,4)

6 6 2 2(6,2) (6,2)


A (faulty) proof tree in natural semantics:(what is missing?)

3 3 5 5

(3,5) (3,5)

3 5 8 4 2 2

4 4 2 2

(4,2) (4,2)

(3 5)*(4 2) 16 * *


Properties of natural semantics

• For each value V,

• For each E, if then E’ is a value (The value variables in the rules are not needed!)

• For each E, there is at most one proof tree, hence

at most one value(main idea of proof: For each expression category, except if, a unique rule.)

Natural semantics is a functional relation --- it defines a partial function

V V

E E


On run-time errors

High level pl’s do not allow run-time errors to occur w/o taking notice:

prevent, detect and treat

A semantics should specify explicitly both

• what are errors,

• error handling


A computation from E may get stuck in a

non-value final expression

3/0, true + 2, if 3 4 5, 2 3

We shall:• Determine the forms of error expressions• Introduce terminology• Add rules that specify error treatment to all specs


Candidate for a reduction: a expression of the forms• •

For both, proper rules certainly do not apply

If there is no transition: a run-time error

4/0, 4 2, false + 3, + (2, 3, 4)(All, except the first, are run-time type errors)

Candidate for a reduction:

either a redex or a run-time error Claim: there is a at most one transition from a

candidate for a reduction (for both , d).

1 2v v

2 3f i v e e


The run-time errors:

3/0, 2+false, + (2,3,4)

2 4

if 5 4 6

1 2 1 , where op v v v

2 2 o , where ( ) is undefinedov f v

if 2 3 where boolv e e v


For each composite expression:

each selected sub-expression

is a candidate for a reduction

Def:• E contains an error:

a selected sub-expression is a run-time error• E is an error: ( a stuck expression)

all its selected sub-expressions are run-time errors

Discuss differences between , d on these concepts


Error treatment --- error rules

We add a special “constant” ER (not a regular expression/value)

We add error rules to specify error handling

An expression may transform in one step to ER iff • It is a run-time error (error axioms)• It contains an error (error propagation rules)

Def: Exval - extended values: Val {ER}

extended expressions: Exp {ER}


Need to change assumption about built-in:

For each built-in operation o, is a total function from Val to Exval (returns a value or ER)

Takes care of

of

where ( ) is undefinedoo v f v


Some error rules:

For other constructs –left to you

For d -- left for you

The new set of rules

Is there a change in selected sub-expressions?

Can rules be improved upon by using v instead of e?

1

1 2

(apply-error) v op

v e

ER

1

1 2

(rator-error) e

e e

ER

ER

2

1 2

(rand-error) e

e e

ER

ER

,ER[ ] ( [ ] )S EL S EL


Example:

ER

(3 5)*(4 5)

applic

* tuple

applic applic

3 5

4 5

seq

(error-rand)

(error-tuple) (2,1)

(apply- error)


Rules for natural semantics:

The other constructs – left to you

The new set:

1

1 2

(nat-applic-er1) v op

v e

ER

1

1 2

(nat-applic-er2) e

e e

ER

ER

2

1 2

(nat-applic-er3) e

e e

ER

ER

,ER[ ] ( [ ] )S EL S EL


Additional properties for semantics with error rules:• Soundness: an expression cannot evaluate to both a

value and ER• Completeness/Progress: If E is not in Exval, then E

evaluates to some XV in Exval (progress --- evaluation does not get stuck)

Determinism implies soundness(we argue that is deterministic, as before)

Completeness: Argue that a candidate for a reduction is never stuck, extend to all expressions

, ,ER[ ] dS EL

Documents

Winter 2004/5Pls – expresssion-lang – Catriel Beeri22 Informal Analysis Goal: prove the semantics is deterministic Stages: Values --- we know they have