CS 655: Programming Languages

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CS 655 Programming Languages
Lecture 2 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 thats 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

• David Evans
• Office 236A, 982-2218
• evans_at_cs.virginia.edu
• http//www.cs.virginia.edu/evans

University of Virginia Computer Science
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Menu

• Survey Summary
• Position Papers Results
• Intro to Formal Semantics
• Operational Semantics

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Students Experience

• C - 92.5 years, C - 78, BASIC 64,
• Pascal 45, Java - 26.5, assemblies 26,
  FORTRAN 22, LISP 17, Perl 14.5
• Ada, ADAMS, Common LISP, Delphi, Elf, Forth,
  Haskell, HP Calculator, Hypertalk, JavaScript
  (large system for DOD!), ksh, LOGO, ML, nawk,
  ObjectPal, Python, Prolog, REBOL, Relay Ladder
  Logic, Sather, Scheme, Smalltalk, SQL, Tcl/Tk

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Position Paper 1

• No ?, 5 ?2 ( 1.4)
• Our not favorite languages
• BASIC (1) C (8 - 1)
• C (3) FORTRAN (2)
• Java (8 - 2) Perl (2 - 2)
• Good papers had to get at underlying issues, not
  just symptoms
• Some misconceptions about GC

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Project Teams

• You have until noon Monday to submit team
  requests
• Use the discussion board if you are looking for
  teammates
• (link off cs655 page)

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Formal Semantics
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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

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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, 3.1.2.6

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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.

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A Toy Language BARK
(Beginners All-Purpose Register Kode)

• Program Instruction Program is a sequence
  of instructions
• Instructions are numbered from 0.
• Instruction
• STORE Loc Literal Loc gets the value of
  Literal
• ADD Loc1 Loc2 Loc1 gets the value of Loc1
  Loc2
• MUL Loc1 Loc2 Loc1 gets the value of Loc1
  Loc2
• HALT Stop program execution, answer in R0
• ERROR Stop program execution, error
• GOTO Loc Jump to instruction corresponding
  to
• value in Loc.
• IF Loc1 THEN Loc1 If value in Loc1 is
  non-zero,
• jump to instruction corresponding
• to value in Loc2.
• Loc R-?0-90-9
• Literal -?0-90-9

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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

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Operational Semantics Game
Abstract Machine
Real World
Program
Initial Configuration
Input Function
Intermediate Configuration
Transition Rules
Intermediate Configuration
Answer
Final Configuration
Output Function
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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
• O F ? Answer (output function)

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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-1
Register-1
Instruction0
Register0
Instruction1
PC
Register1
Instruction2
Register2
.
.
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Input Function I Program ? C

• C Instructions x PC x RegisterFile where
• For a Program with n instructions numbered from 0
  to n - 1Instructionsm Programm for m gt
  0 m lt n
• Instructionsm ERROR otherwise
• PC 0
• RegisterFilen 0 for all integers n

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Output Function
Final Configurations
F Instructions x PC x RegisterFile where
InstructionsPC HALT

• OF ? Answer
• Answer value in RegisterFile0

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Operational Semantics Game
Abstract Machine
Real World
Program
Initial Configuration
Input Function
Intermediate Configuration
Transition Rules
Intermediate Configuration
Answer
Final Configuration
Output Function
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Form of Transition Rules

• Antecedents

c ? c
Where c is a member of C .
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STORE Loc Literal

• InstructionsPC STORE Loc Literal

lt Instructions x PC x RegisterFile gt ? lt
Instructions x PC x RegisterFile gt where PC
PC 1 RegisterFilen RegisterFilen if n
? Loc RegisterFilen value of Literal if n
? Loc
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ADD Loc1 Loc2

• InstructionsPC ADD Loc1 Loc2

lt Instructions x PC x RegisterFile gt ? lt
Instructions x PC x RegisterFile gt where PC
PC 1 RegisterFilen RegisterFilen if n
? Loc RegisterFilen RegisterFileLoc2 if
n ? Loc1
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GOTO Loc

• InstructionsPC GOTO Loc

lt Instructions x PC x RegisterFile gt ? lt
Instructions x PC x RegisterFile gt where PC
value in RegisterFileLoc
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IF Loc1 THEN Loc2

• InstructionsPC IF Loc1 THEN Loc2

lt Instructions x PC x RegisterFile gt ? lt
Instructions x PC x RegisterFile gt where PC
value in RegisterFileLoc2 if Loc1 is
non-zero PC PC 1 otherwise
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Whats it good for?

• Understanding programs
• Write a compiler or interpreter (?)
• Prove properties about programs and languages

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Variation BARK-forward

• Same as BARK except
• GOTO Loc Jump forward Loc instructions. Loc
  must
• be positive.
• IF Loc1 THEN Loc2 If value in Loc1 is
  non-zero, jump forward
• value in Loc2. instructions. Loc2 must
  be positive.

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GOTO Loc
BARK

• InstructionsPC GOTO Loc

lt Instructions x PC x RegisterFile gt ? lt
Instructions x PC x RegisterFile gt where PC
value in RegisterFileLoc
InstructionsPC GOTO Loc,value in Loc gt 0
lt Instructions x PC x RegisterFile gt ? lt
Instructions x PC x RegisterFile gt where PC
PC value in RegisterFileLoc
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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

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Energy Function

• Energy C ? positive integer
• C lt Instructions x PC x RegisterFile gt
• Energy h PC where
• h is an integer such that
• Instructionsk error for all k gt h

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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
- 1Instructionsm Programm for m gt 0
m lt n Instructionsm ERROR otherwise PC
0 RegisterFilen 0 for all integers n
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Initial Configuration Energy

• Energy (C ) n
• where n is number of program instructions
• PC 0
• Instructionm ERROR for m gt n

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Energy 0 ? Terminated

• Energy h PC where
• h is an integer such that
• Instructionsk error for all k gt h
• so, Energy 0 ?
• h PC and
• InstructionsPC ERROR
• No transitions for configuration where
  InstructionsPC ERROR

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STORE reduces Energy
InstructionsPC STORE Loc Literal
lt Instructions x PC x RegisterFile gt ? lt
Instructions x PC x RegisterFile gt where PC
PC 1 RegisterFilen RegisterFilen if n
? Loc RegisterFilen value of Literal if n
? Loc
Energy (ltInstructions x PC x RegisterFilegt) h
PC Energy (ltInstructions x PC x RegisterFilegt)
h (PC 1) h depends only on Instructions,
doesnt change Therefore Energy lt Energy
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GOTO reduces Energy
InstructionsPC GOTO Loc,value in Loc gt 0
lt Instructions x PC x RegisterFile gt ? lt
Instructions x PC x RegisterFile gt where PC
PC value in RegisterFileLoc
Energy(ltInstructions x PC x RegisterFilegt) h -
PC Energy(ltInstructions x PC x RegisterFilegt)
h (PC RegisterFileLoc) but antecedent
says RegisterFileLoc gt 0, so PC
RegisterFileLoc gt PC and Energy lt Energy.
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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)

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Charge

• Problem Set 1
• Develop Operational Semantics for simplified
  PostFix using RVM
• Prove termination property
• Final project team requests by Monday
• Next time
• Projects
• Discuss Wenger paper come to class prepared to
  discuss how well his milestones work today