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David Evans CS655: Programming Languages University of Virginia Computer Science Lecture 13: Operational Semantics “Then.

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Presentation on theme: "David Evans CS655: Programming Languages University of Virginia Computer Science Lecture 13: Operational Semantics “Then."— Presentation transcript:

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2 David Evans CS655: Programming Languages University of Virginia Computer Science Lecture 13: Operational Semantics “Then you should say what you mean,” the March Hare went on. “I do,” Alice hastily replied; “at least – at least I mean what I say – that’s the same thing, you know.” “Not the same thing a bit!” said the Hatter. “Why, you might as well say that ‘I see what I eat’ is the same thing as ‘I eat what I see’!" Lewis Carrol, Alice in Wonderland “When I use a word,” Humpty Dumpty said, in a rather scornful tone, “it means just what I choose it to mean – neither more nor less.” Lewis Carroll, Through the Looking Glass

3 1 March 2001CS 655: Lecture132 Menu Quiz Results Intro to Formal Semantics Operational Semantics

4 1 March 2001CS 655: Lecture133 Quiz Results Preferences –Mostly formal techniquesx –Balancedxx –Mostly language designxxxxx –Just language designxx Mock Trial –Heard aboutYes6No3 Yes, very 1 Yes 2 Maybe 1.5 No, research0.5 No, sleeping 1 Speed: way too fast (0), too fast (6), wrote in just right/both (3), too slow (1)

5 1 March 2001CS 655: Lecture134 Other Comments “disappointed to hear that there were no more programming assignments...I’d be interested in playing with and analyzing languages like Smalltalk, CLU, and APL...” “though I’ve found a good deal of the material interesting, I don’t necessarily feel entirely prepared to write a Why X is not my favorite Programming Language” “I’m tired of Scheme...mostly because the MIT scheme interpreter is poorly written and doesn’t allow you to easily perform editing operations...” Run M-x scheme in emacs

6 1 March 2001CS 655: Lecture135 Other Comments “complaint is that we’re doing homework on old stuff that is no longer focused on in class, making it difficult for me to devote my energies to the new stuff.” “hasn’t taken the approach I would’ve expected – such as covering PL concepts (types, binding, etc.) in the context of a range of languages” “hope we can get lectures before class, so that we can have some preparation before class,..., and take some notes beside the lectures.” “there are too many readings. I don’t have enough time to read them all.” (also checked “Mostly language design and history”)

7 1 March 2001CS 655: Lecture136 Find a Lambda calculus term that has a normal form, but that will not be reduced to that form if beta reductions are not done in normal order. Like asking: “Find a Scheme program that terminates with lazy evaluation, but doesn’t terminate with eager evaluation. Translate it into Lambda calculus.”

8 1 March 2001CS 655: Lecture137 What does a program mean? Compile and run –Implementation dependencies –Not useful for reasoning Informal Semantics –Natural language description of PL Formal Semantics –Description in terms of notation with formally understood meaning

9 1 March 2001CS 655: Lecture138 Why not informal semantics? Two types have compatible types if their types are the same [footnote: Two types need not be identical to be compatible.]. ANSI C Standard,

10 1 March 2001CS 655: Lecture139 Formal Semantics Approaches Operational –Map to execution of a virtual machine –Easier to understand, harder to reason about –Depends on informal understanding of machine Denotational –Map parts to mathematical meanings, rules for composing meanings of pieces –Harder to understand, easier to reason about –Depends on informal understanding of mathematics Lots of others: Algebraic, Translator, etc. What about Lambda Calculus?

11 1 March 2001CS 655: Lecture1310 A Toy Language: BARK Program ::= Instruction*Program is a sequence of instructions Instructions are numbered from 0. Instruction ::= STORE Loc Literal Loc gets the value of Literal | ADD Loc 1 Loc 2 Loc 1 gets the value of Loc 1 + Loc 2 | MUL Loc 1 Loc 2 Loc 1 gets the value of Loc 1 * Loc 2 | HALT Stop program execution, answer in R0 | ERROR Stop program execution, error | GOTO LocJump to instruction corresponding to value in Loc. | IF Loc 1 THEN Loc 1 If value in Loc 1 is non-zero, jump to instruction corresponding to value in Loc 2. Loc ::= R[ - ] ? [ ][ ]* Literal ::= [ - ] ? [ ][ ]* (Beginner’s All-Purpose Register Kode)

12 1 March 2001CS 655: Lecture1311 A BARK Program [0]STORE R0 1 [1]STORE R1 10 [3]STORE R2 –1 [4] STORE R3 6 [5] STORE R4 8 [6]IF R1 THEN R4 [7]HALT [8] MUL R0 R1 [9]ADD R1 R2 [10]GOTO R3

13 1 March 2001CS 655: Lecture1312 Operational Semantics Game Input Function Abstract Machine Initial Configuration Final Configuration Output Function Answer Intermediate Configuration Intermediate Configuration Transition Rules Real World Program

14 1 March 2001CS 655: Lecture1313 Structured Operational Semantics SOS for a language is five-tuple: C Set of configurations for an abstract machine  Transition relation (subset of C x C ) I Program  C (input function) F Set of final configurations OF  Answer(output function)

15 1 March 2001CS 655: Lecture1314 Abstract Machine: Register Virtual Machine (RVM) Configuration defined by: –Array of Instructions –Program counter –Values in registers (any integer) C = Instructions x PC x RegisterFile Instruction[0] Instruction[1] Instruction[2] …. Instruction[-1] …. PC Register[0] Register[1] Register[2] …. Register[-1] ….

16 1 March 2001CS 655: Lecture1315 Input Function: I : Program  C C = Instructions x PC x RegisterFile where For a Program with n instructions numbered from 0 to n - 1: Instructions[m] = Program[m] for m >= 0 && m < n Instructions[m] = ERROR otherwise PC = 0 RegisterFile[n] = 0 for all integers n

17 1 March 2001CS 655: Lecture1316 Final Configurations F = Instructions x PC x RegisterFile where Instructions[PC] = HALT Output Function O : F  Answer Answer = value in RegisterFile[0]

18 1 March 2001CS 655: Lecture1317 Operational Semantics Game Input Function Abstract Machine Initial Configuration Final Configuration Output Function Answer Intermediate Configuration Intermediate Configuration Transition Rules Real World Program

19 1 March 2001CS 655: Lecture1318 Form of Transition Rules Antecedents c  c ’ Where c is a member of C.

20 1 March 2001CS 655: Lecture1319 STORE Loc Literal Instructions[PC] = STORE Loc Literal  where PC’ = PC + 1 RegisterFile’[n] = RegisterFile[n] if n  Loc RegisterFile’[n] = value of Literal if n  Loc

21 1 March 2001CS 655: Lecture1320 ADD Loc 1 Loc 2 Instructions[PC] = ADD Loc 1 Loc 2  where PC’ = PC + 1 RegisterFile’[n] = RegisterFile[n] if n  Loc RegisterFile’[n] = RegisterFile[Loc 2 ] if n  Loc 1

22 1 March 2001CS 655: Lecture1321 GOTO Loc Instructions[PC] = GOTO Loc  where PC’ = value in RegisterFile[Loc]

23 1 March 2001CS 655: Lecture1322 IF Loc 1 THEN Loc 2 Instructions[PC] = IF Loc 1 THEN Loc 2  where PC’ = value in RegisterFile[Loc 2 ] if Loc 1 is non-zero PC’ = PC + 1 otherwise

24 1 March 2001CS 655: Lecture1323 What’s it good for? Understanding programs Write a compiler or interpreter (?) Prove properties about programs and languages

25 1 March 2001CS 655: Lecture1324 Variation: BARK-forward Same as BARK except: GOTO LocJump forward Loc instructions. Loc must be positive. | IF Loc 1 THEN Loc 2 If value in Loc 1 is non-zero, jump forward value in Loc 2. instructions. Loc 2 must be positive.

26 1 March 2001CS 655: Lecture1325 GOTO Loc Instructions[PC] = GOTO Loc  where PC’ = value in RegisterFile[Loc] BARK: Instructions[PC] = GOTO Loc,value in Loc > 0  where PC’ = PC + value in RegisterFile[Loc]

27 1 March 2001CS 655: Lecture1326 Proving Termination Idea: Prove by Induction –Define a function Energy: C  positive integer –Show Energy of all Initial Configurations is finite –Show there are no transitions from a configuration with Energy = 0 –If C  C’ is a valid transition step, Energy of C’ must be less than Energy of C

28 1 March 2001CS 655: Lecture1327 Energy Function Energy: C  positive integer C = Energy = h – PC where h is an integer such that Instructions[k] = error for all k >= h

29 1 March 2001CS 655: Lecture1328 Initial Configurations For all Initial Configurations C, Energy is finite. Consider Input Function: C = Instructions x PC x RegisterFile where For a Program with n instructions numbered from 0 to n - 1: Instructions[m] = Program[m] for m >= 0 && m < n Instructions[m] = ERROR otherwise PC = 0 RegisterFile[n] = 0 for all integers n

30 1 March 2001CS 655: Lecture1329 Initial Configuration Energy Energy ( C ) = n where n is number of program instructions PC = 0 Instruction[m] = ERROR for m >= n

31 1 March 2001CS 655: Lecture1330 Energy = 0  Terminated Energy = h – PC where h is an integer such that Instructions[k] = error for all k >= h so, Energy = 0  h = PC and Instructions[PC] = ERROR No transitions for configuration where Instructions[PC] = ERROR

32 1 March 2001CS 655: Lecture1331 STORE reduces Energy Instructions[PC] = STORE Loc Literal  where PC’ = PC + 1 RegisterFile’[n] = RegisterFile[n] if n  Loc RegisterFile’[n] = value of Literal if n  Loc Energy ( ) =h – PC Energy ( ) = h – (PC + 1) h depends only on Instructions, doesn’t change Therefore: Energy’ < Energy

33 1 March 2001CS 655: Lecture1332 GOTO reduces Energy Instructions[PC] = GOTO Loc,value in Loc > 0  where PC’ = PC + value in RegisterFile[Loc] Energy( ) = h - PC Energy( ) = h – (PC + RegisterFile[Loc]) but antecedent says RegisterFile[Loc] > 0, so PC + RegisterFile[Loc] > PC and Energy’ < Energy.

34 1 March 2001CS 655: Lecture1333 To complete proof… Show the same for every transition rule. Then: –Start with finite energy, –Every transition reduces energy, –Energy must eventually reach 0. –And energy 0 means we terminated. Minor flaw? could skip 0 (e.g., Energy = –1)

35 1 March 2001CS 655: Lecture1334 Charge Gifford’s Ch 3 uses a stack-based virtual machine to provide an operational semantics for a stack-based language Next time: –Projects Kickoff – start thinking about things you want to do –Types and Static Semantics


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