1 semantics q1 2007 s emantics (q1,’07) week 2 jacob andersen phd student andersen@daimi.au.dk

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1

Semantics Q1 2007

SEMANTICS (Q1,’07)Week 2

Jacob AndersenPhD student

andersen@daimi.au.dk

2

Semantics Q1 2007

Week 2 - Outline• Highlights from last week• Syntax vs. Semantics• Virtual Machine Semantics• Structural Operational Semantics

• Expressions (Exp)» Big-step vs. small-step semantics» Side-effects» Behavior and Equivalence, …

• Boolean Expressions (BExp)» Lazy evaluation, …

• Commands (Com)» …

3

Semantics Q1 2007

HIGHLIGHTS FROM LAST WEEK

4

Semantics Q1 2007

• Correlation:• Investment ~ Benefit

Ex: Studying ~ Exam result

• Do Active Learning:• Study = read + make exercises

+ reflect (for your own sake)!

Active vs. Passive Learning

Active Learning Passive Learning

vs.

Source: J. Biggs, “Teaching for quality learning at university”

5

Semantics Q1 2007

Why bother learning Semantics?!?

Why Semantics?!?

Fully understand

a prog. lang.

Semantics-preserving

prog. changes

designnewprog. lang.s

rapidly

learn new

prog. lang.debuggingprograms

cross-

compilationautomatedoptimization

automatederror detection

writefaster

programs

write"better"

programs

$$

$$$$

Pass thiscourse

6

Semantics Q1 2007

Relations• A k-ary relation, L, over the sets

S1, S2, … , Sk

is a subsetL S1 S2 … Sk

7

Semantics Q1 2007

Inference Systems• Inference System:

• Inductive (recursive) specification of relations• Consists of axioms and rules

|_even n

|_even m

m = n+2

premise(s)

conclusion

side-condition(s)

Note: The “ “ is actually just another way of writing a “” !

8

Semantics Q1 2007

Transition Systems• A Transition System is a structure:

is a set of configurations is a binary relation

(called the transition relation)

• Alternative flavours:– Labelled TS– Terminal TS– Etc…

,

, , T where T and t T : : t

, A , and A

9

Semantics Q1 2007

SYNTAX VS SEMANTICS

10

Semantics Q1 2007

Syntax and Semantics• Syntax:

• = Structure

– vs.

• Semantics:• = Meaning

"Eats shoots and leaves"

"Eats shoots and leaves"

"Colorless Green Ideas Sleep Furiously"-- N. Chomsky (1957)

; ;

Has legal syntactic structure, but no meaning

( )

, ,

11

Semantics Q1 2007

Syntax vs. Semantics (cont'd)• In natural languages (English, Danish, ...):

• Syntax Semantics–

• In programming languages (C, Java, ML, ...):• Syntax Semantics

» By design (for very good reasons...):

"Time flies like an arrow "

"Fruit flies like a banana "

---------N --------V -----------N-------P

-------------------N -------V ---------------N

But, if I replace

two nouns(syntactic)

replace noun replace noun

12

Semantics Q1 2007

Level of Abstraction• Semantic specification(s):

• Appropriate level of abstraction:» ...so that we can use the semantics for something !!!

• See behind concrete details;– Perceive only what is relevant and at the appropriate

level of abstraction)!

is_busy is_busy is_done

high

er le

vel o

f ab

stra

ctio

n

..., transistor57824 = 5V , transistor57825 = 0V , ...

..., transistor57824 = 5V , transistor57825 = 5V , ...

zzz

13

Semantics Q1 2007

THE LANGUAGE “L”

14

Semantics Q1 2007

The Language ”L”• Basic Syntactic Sets:

• Truthvalues:ranged over by: t , t’, t0, …

• Numbers:ranged over by: m, n, …

• Variables:ranged over by: v, v’, …

T = {tt, ff}

N = {0, 1, 2, …}

VAR = {a, b, c, …, z}

Meta-variables

15

Semantics Q1 2007

The Language ”L”• Derived Syntactic Sets:

• Arithmetic Expressions (e Exp):

• Boolean Expressions (b BExp):

• Commands (c Com):

e ::= n | v | e + e’ | e – e’ | e e’

b ::= t | e = e’ | b or b’ | ~ b

c ::= nil | v := e | c ; c’ | if b then c else c’ | while b do c

i.e. Exp = “all ASTs generated by this grammar”

16

Semantics Q1 2007

VIRTUAL MACHINE SEMANTICS

'The world before 1981'

17

Semantics Q1 2007

• Value Stack:– Set ranged over by: S

• Memories:– Functions ranged over by: M

• Control Stack:– Set ranged over by: C

• SMC Machine (Trans. Sys. Semantics):– Configurations:

Virtual Machines (SMC Expressions)

= Value Stack Memories Control Stack

( T N Var BExp Com )*

VAR N

( Exp { +, –, } ...)*

Initially commands/expressions, and later bits of commands/expr's

For accumulating partial resultsand other pieces needed

18

Semantics Q1 2007

• Configurations:

• Notation: , M VAR N

– i.e., “Memory Update”

Notation: Memory Update

:=

( T N ... )* VAR N ( Exp { +, –, } ...)*

m , if v = v’M’(v’) = M(v’) , otherwise

M[m/v]

M[m/v] = M’ where

19

Semantics Q1 2007

Virtual Machine Transitions• Configurations:

• Transitions:• Defined "by case" according to top of control stack:

[case n]: < S , M , n C > < n S , M , C >

[case v]: < S , M , v C > < M(v) S , M , C >

[case ee’]: < S , M , e e’ C > < S , M , e e’ C >

[case ]: < m’ m S , M , C > < n S , M , C > …where n = m m’

Syntactic ‘+’

Semantic ‘+’

:=

( T N ... )* VAR N ( Exp { +, –, } ...)*

20

Semantics Q1 2007

Virtual Machine: Example• Given program: and memory:(((x + 1) – y) 7)

< , M, (((x + 1) – y) 7) >

< , M, (x + 1) ; y ; – ; 7 ; >

< , M, ((x + 1) – y) ; 7 ; >

M = [x=9,y=4]

< , M, x ; 1 ; + ; y ; – ; 7 ; >

< , M, ((x + 1) – y) ; 7 ; > < , M, (x + 1) ; y ; – ; 7 ; >

< , M, x ; 1 ; + ; y ; – ; 7 ; >

< 9, M, 1 ; + ; y ; – ; 7 ; >

< 9, M, 1 ; + ; y ; – ; 7 ; > < 1 ; 9, M, + ; y ; – ; 7 ; >

< 9 ; 1, M, + ; y ; – ; 7 ; > < 10, M, y ; – ; 7 ; >

< 10, M, y ; – ; 7 ; > < 4 ; 10, M, – ; 7 ; >

< 4 ; 10, M, – ; 7 ; > < 6, M, 7 ; >

< 6, M, 7 ; > < 7 ; 6, M, >

< 7 ; 6, M, > < 42, M, >

21

Semantics Q1 2007

VM Semantics: Major Drawbacks!

• Advantage:• Easy to implement (and efficient)

• Drawbacks:• “High-level language understood in terms of low-level

machine code”• Non-intuitive • Too concrete (e.g., stack for computation fragments) • Indirect semantics (not syntax directed) • Computational step?

“Many other machine along these lines […]. They all have a tendency to pull the syntax into pieces or at any rate to wander around the syntax creating various complex symbolic structures which do not seem particularly forced by the demands of the language itself”

- Gordon Plotkin, ‘81

22

Semantics Q1 2007

ARITHMETIC EXPRESSIONS

23

Semantics Q1 2007

• Given program: and memory:

VM: Processing of Additions

< , M, ((1 + (2 + 3)) + 4) > < , M, (1 + (2 + 3)) 4 + >

< , M, (1 + (2 + 3)) 4 + > < , M, 1 (2 + 3) + 4 + >

< , M, 1 (2 + 3) + 4 + > < 1, M, (2 + 3) + 4 + >

< 1, M, (2 + 3) + 4 + > < 1, M, 2 3 + + 4 + >

< 1, M, 2 3 + + 4 + > < 2 1, M, 3 + + 4 + >

< 2 1, M, 3 + + 4 + > < 3 2 1, M, + + 4 + >

< 3 2 1, M, + + 4 + > < 5 1, M, + 4 + >

< 5 1, M, + 4 + > < 6, M, 4 + >

< 6, M, 4 + > < 4 6, M, + >

< 4 6, M, + > < 10, M, >

((1 + (2 + 3)) + 4) M

rearrange

rearrange

addition1!

addition2!

rearrange

addition3!

rearrange

rearrange

rearrange

rearrange

12 3

Note: Only three transitions are of real interest as “system events”

24

Semantics Q1 2007

• Ideally we would like:

• i.e., the following transition sequence:

Ideal Processing of Additions

((1 + (2 + 3)) + 4) ((1 + 5) + 4)

explanation

((1 + 5) + 4)

explanation

(6 + 4)

(6 + 4)

explanation

10

((1 + (2 + 3)) + 4)

explanation

((1 + 5) + 4)

explanation

(6 + 4)

explanation

10

(aka. derivation sequence)(aka. reduction sequence)

25

Semantics Q1 2007

Informal vs. Formal Specification

• Informally:• “One evaluates from left to right...”

• Description (pseudo-formally):• CONSTANTS:

– Any constant, n, is already evaluated (with itself as value)

• SUMS:– Step 1. Evaluate to obtain result ;– Step 2. Evaluate to obtain result ;– Step 3. Add and to obtain final result .

n

e0 + e1

e0 n0

e1 n1

n0 n1 m

26

Semantics Q1 2007

Inference System Semantics

• Inference System Semantics:• Abbreviate as

» Meaning: “e terminates and evaluates to v”

e v

‘L’ Exp N

(e,v) ‘L’

Q: Did I just solve the halting problem here?!?

A: No; nobody can decide “e v” automatically!

A: Actually, for Exp termination is decidable…

27

Semantics Q1 2007

Inference System Sem. (cont’d)

• Inference System Semantics:

e0 + e1 mm = n0 + n1

e0 n0 e1 n1[SUM][CON] n n

‘L’ Exp N

Syntactic ‘+’Semantic ‘+’

28

Semantics Q1 2007

Example Revisited (big-step)

• Inference Tree:

((1 + (2 + 3)) + 4) 10

(1 + (2 + 3)) 6 4 4

1 1 (2 + 3) 5

2 2 3 3

e0 + e1 mm = n0 + n1

e0 n0 e1 n1[SUM][CON] n n

[SUM]

[CON]

[CON]

[SUM]

[SUM]

[CON] [CON]“big-step semantics”

“small-step semantics”vs.

29

Semantics Q1 2007

Small-step Description• Description:

• CONSTANTS:– Any constant, n, is already evaluated (with itself as value)

• SUMS:– Step 1. If is not a constant,

evaluate it (one step) to obtain result ;– Step 2. If is a constant, but is not,

evaluate it (one step) to obtain result ;– Step 3. If and are both constants,

add them to obtain result, say .

n

e0 + e1

e0

e0’

e0 e1

e1’

e0 e1

m

30

Semantics Q1 2007

Small-step Formalization

• Transition System Semantics:• Configurations:

• Final Configurations:

• Transition Relation:– Abbreviate as

» Meaning: “e evaluates to e’ in one step”

L := Exp

TL := N Exp

L Exp Exp

e e’(e,e’) ‘L’

31

Semantics Q1 2007

Small-step Semantics

• Transition System Semantics:

L := Exp

TL := N Exp

L Exp Exp

[SUM1]

e0 + e1 e0’ + e1

e0 e0’

[SUM2]

n0 + e1 n0 + e1’

e1 e1’

[SUM3]

n0 + n1 mm = n0 + n1

We call this a STRUCTURAL OPERATIONAL SEMANTICS

32

Semantics Q1 2007

Example Revisited (small-step)

• Transition sequence (with explanation):

((1 + (2 + 3)) + 4) ((1 + 5) + 4)

(1 + (2 + 3)) (1 + 5)

(2 + 3) 5

[SUM1]

[SUM2]

[SUM3]

“order of discovery”

find predicates make conclusions

33

Semantics Q1 2007

Transition Sequence Information

• A transition sequence

…specifies two things:

• 1. A “sequence of steps” themselves (here additions)

• 2. …and reasons why they should be performed

e e’ e’’ … m

((1 + (2 + 3)) + 4)

explanation

((1 + 5) + 4)

34

Semantics Q1 2007

Adding Variables• Language L’:

– Arithmetic Expressions (e Exp)•

• Structural Operational Semantics:– Configurations:

– Final Configurations:–

– Transition Relation…–

e ::= n | v | e0 + e1

L’ := Exp Store

TL’ := N Store L’

Note the change in terminology: Store <--- Memory <--- MThe term Store is more generally accepted (as an abstraction of a computer’s physical memory)

Store = Var N (= Memory)where

A configuration now looks like: <e,><e,> L <e’,'>

35

Semantics Q1 2007

Small-step Semantics w/ Stores• Structural Operational Semantics (w/ Stores):

[SUM1]< e0 + e1 , > < e0’ + e1 , ’ >

< e0 , > < e0’ , ’ >

[SUM2]

[SUM3]< n0 + n1 , > < m , >

m = n0 + n1

< n0 + e1 , > < n0 + e1’ , ’ >< e1 , > < e1’ , ’ >

< v , > < m , >[VAR]

Store = Var Nm = (v)

36

Semantics Q1 2007

In fact: we have no Side-Effects!

• No side-effects! •

[SUM1]< e0 + e1 , > < e0’ + e1 , ’ >

< e0 , > < e0’ , ’ >

[SUM2]

[SUM3]< n0 + n1 , > < m , >

m = n0 + n1

< n0 + e1 , > < n0 + e1’ , ’ >< e1 , > < e1’ , ’ >

< v , > < m , >[VAR]

Store = Var Nm = (v)

e,: <e,> <e’,’> => = ’Easily proved by structural induction [later…]

37

Semantics Q1 2007

Explicit Absence of Side-effects• SOS Semantics:

[SUM1]

|_ e0 + e1 e0’ + e1

|_ e0 e0’

[SUM2]

[SUM3] |_ n0 + n1 m

m = n0 + n1

|_ n0 + e1 n0 + e1’

|_ e1 e1’

|_ v m

[VAR]Store = Var Nm = (v)

|_ e e’ <e,> <e’,>

Note: Absence of side-effects is now explicit

for

38

Semantics Q1 2007

Explicit Absence of Side-effects• Terminology (more common):

[SUM1]

|_ e0 + e1 e0’ + e1

|_ e0 e0’

[SUM2]

[SUM3] |_ n0 + n1 m

m = n0 + n1

|_ n0 + e1 n0 + e1’

|_ e1 e1’

|_ v m

[VAR]

Env = Var N (= Store)m = (v)

Note the change in terminology: Env <--- Store <---

39

Semantics Q1 2007

Exp: Behavior and Equivalence• Definitions:

• DEF: "Behavior" (eval):

» Note: eval is only a partial function(e.g., if we have division; div-by-zero)

• DEF: "Program equivalence" (''):

» Examples:

eval(e,) = m <=>def <e,> E* <m,>

e e’ <=>def : eval(e,) = eval(e’,)

1+1 2 x+y y+x z+z 2*z

Live Exercise: Does "" depend on the semantics (if so, how/where)?

40

Semantics Q1 2007

BOOLEAN EXPRESSIONS

41

Semantics Q1 2007

Boolean Expressions BExp• Language B:

• Boolean Expressions (b BExp):–

• Structural Operational Semantics:• Configurations:

– where

• Final Configurations:–

• Transition Relation…– short-hand for

b ::= t | e = e’ | b or b’ | ~ b

B := BExp Env

TB := T Env B

Env = Var N

|_ b B b’

We could also have used a Store (we actually have to if BExp had had side-effects)

((b,),(b’,)) B

42

Semantics Q1 2007

Boolean Expressions BExp (cont’d)

• Structural Operational Semantics:•

• Note that:– The (this) boolean expression transition system...:

– …uses the arithmetic expression transition system…:

– ...and does it transitively (E*) in “one step”!

|_ e0 = e1 B t

[EQ]B |_ e0 E

* n0 |_ e1 E* n1

B , TB , B

E , TE , E

t = tt, n0 = n1

ff, n0 n1

See [Plotkin, p. 42-44] for alternative semantics (with a smaller step size)

43

Semantics Q1 2007

Disjunction: “or”• SOS for Disjunction (“or”):

• What about the boolean expression: ...?!?

|_ b0 or b1 B b0’ or b1

|_ b0 B b0’[OR1]B

|_ t0 or b1 B t0 or b1’

|_ b1 B b1’[OR2]B

|_ t0 or t1 B t

[OR3]B t = t0 t1

"tt or b"

Live Exercise:syntax vs semanticoperators?

44

Semantics Q1 2007

Disjunction: Lazy Evaluation• Lazy Semantics for Disjunction (“or”):

|_ b0 or b1 L b0’ or b1

|_ b0 L b0’[OR1]L

[OR2]L

|_ ff or b1 L b1

[OR3]L

b: tt or b tt |_ tt or b1 L tt...exploiting:

b: ff or b b

...exploiting:

45

Semantics Q1 2007

Eager vs. Lazy Semantics for “or”• Eager Semantics: ‘B’ • Lazy Semantics: ‘L’

• Relationship: ?– Stuck configurations?– …e.g. div-by-zero?– Assignment?– Non-termination?

b := tt or (3 = 1-2)

b,: |_ b B* t <=> |_ b L

* t

b := tt or (1 = 2/0)

b := tt or (x := tt)“side-effects”

b := tt or loop()

46

Semantics Q1 2007

– For the language, BExp, we have that:

– Whereas:

Eager vs. Lazy Semantics (cont’d)

b,: <b,> B* <t,’> ==> <b,> L

* <t,”>

b,: <b,> B* <t,’> <== <b,> L

* <t,”>

Can be proved by structural induction [later…]

Easily disproved by a counterexample (e.g. )

• Relationship: ?– Stuck configurations?– …e.g. div-by-zero?– Assignment?– Non-termination?

b := tt or (3 = 1-2)

b,: |_ b B* t <=> |_ b L

* t

b := tt or (1 = 2/0)

b := tt or (x := tt)“side-effects”

b := tt or loop()

b := tt or (3 = 1-2)

47

Semantics Q1 2007

Parallel Evaluation of Disjunction

• Live exercise:• Evaluate BExps non-deterministically (via.

interleaving or in parallel) left / right operands to or:

• [Think 3 mins; then interactively on the whiteboard]

‘P’

b := tt or (3 = 1-2)

b := tt or (1 = 2/0)

b := tt or (x := tt)“side-effects”

b := tt or loop()

• Relationship: ?– Stuck configurations?– …e.g. div-by-zero?– Assignment?– Non-termination?

b,: |_ b B* t <=> |_ b P

* t

48

Semantics Q1 2007

Negation: “~b”• Boolean Expressions (b BExp):

• Live exercise:• [Think 3 mins; then interactively on the whiteboard]

b ::= t | e = e’ | b or b’ | ~ b

49

Semantics Q1 2007

BExp: Behavior and Equivalence

• Definitions:• DEF: "Boolean Expression Behavior":

» Note: eval is only a partial function(e.g., if we have division; div-by-zero)

• DEF: "Boolean Expression Equivalence":

» Examples:

eval(b,) = t <=>def <b,> B* t

b b’ <=>def : eval(b,) = eval(b’,)

b ~~b 1 = 2 ff x = x tt

50

Semantics Q1 2007

COMMANDS

51

Semantics Q1 2007

SOS of Commands• Commands (c Com):

• Structural Operational Semantics:• Configurations:

» where

• Final Configurations:» (?)

• Transition Relation…» for

c ::= nil | v := e | c ; c’ | if b then c else c’ | while b do c

C := (Com Store)

TC C

Store = Var N

<c,> C <c’,’>

Here we need a Store (since we have to model side-effects)

((c,),(c’,’)) C

Problem: how do we end computation?!?

52

Semantics Q1 2007

SOS of Commands (cont’d)• Commands (c Com):

• Structural Operational Semantics:• Configurations:

» where

• Final Configurations:» (!)

• Transition Relation…» for

c ::= nil | v := e | c ; c’ | if b then c else c’ | while b do c

C := (Com Store) Store

TC := Store C

Store = Var N

<c,> C <c’,’> ((c,),(c’,’)) C

<c,> C ’ ((c,),’) C(!)forand

Here we need a Store (since we have to model side-effects)

53

Semantics Q1 2007

Nil and Assignment (SOS)• SOS for nil:

• SOS for assignment:•

v := e

< v := e , > C ’

< e , > A* < m , >

[ASS]C

’ = [m/v]

< nil , > C [NIL]C

54

Semantics Q1 2007

Sequential Composition (SOS)• SOS for Sequential Composition:

• Diagramatically:• “(Control) flow-charts”

c0 ; c1

< c0 ; c1 , > C < c0’ ; c1 , ’ >

< c0 , > C < c0’ , ’ >[SEQ1]C

< c0 ; c1 , > C < c1 , ’ >

< c0 , > C ’[SEQ2]C

c c’

55

Semantics Q1 2007

If-then-else (SOS)• SOS for “if-then-else”:

• Diagramatically:• “(Control) flow-charts”

if b then c else c’

< if b then c else c’ , > C < c , >

< b , > B* < tt , >[IF1]C

< if b then c else c’ , > C < c’ , >

< b , > B* < ff , >[IF2]C

c

c’

b

tt

ff

56

Semantics Q1 2007

While-do (SOS)• SOS for “while-do”:

• Diagramatically:• “(Control) flow-charts”

while b do c

< while b do c, > C < c ; while b do c , >

< b , > B* < tt , >[WH1]C

< while b do c , > C

< b , > B* < ff , >[WH2]C

tt

ff

cb

57

Semantics Q1 2007

Com: Behavior and Equivalence

• Definitions:• DEF: "Command Behavior":

» Note: exec is only a partial function(e.g. non-termination, div-by-zero, …)

• DEF: "Command Equivalence":

» Examples:

exec(c,) = ’ <=>def <c,> C* ’

c c’ <=>def : exec(c,) = exec(c’,)

if b then c else c’ if ~b then c’ else c

while ff do c nil

58

Semantics Q1 2007

"Three minutes paper"• Please spend three minutes writing down the

most important things that you have learned today (now).

After 1 dayAfter 1 week

After 3 weeksAfter 2 weeks

Right away

59

Semantics Q1 2007

Remember to sign up or check that you have been signed up for the exam !!!

Sept. 1 - 15

60

Semantics Q1 2007

Next week: big-step vs. small-step,errors, type checking

Any Questions?

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