Programming Languages

Transcription

1 Programming Languages Jörg Kreiker Chair for Theoretical Computer Science Prof. Esparza TU München winter term 2010/2011

2 3 Lecture 7 Continuations

3 4 Continuations why? Goal 2: To use language features such as higher-order functions and continuations... (today s lecture) Goal 1: To write short error-free programs... use a formal system to verify them... (tutorial next week) Goal 6: To write clear and error-free programs of sufficient complexity (today: regular expression parser)

4 5 Continuations what? Continuation of a computation is what is executed next In C: x = 1; y = 2; continuation after x = 1; is y = 2; In Haskell: id x k = k x; identity in continuation-passing style, where k is the continuation Languages may allow to store the current continuation and to resume it afterwards (first-class continuations) Schmeme: call/cc Haskell: continuation monad SML, Ruby: callcc C: setcontext Perl: Coro and Continuity

5 6 Continuations what? Continuations are a powerful mechanism to explicitly manipulate control-flow Other examples: goto, break, return Cooler examples: fibers (Ruby), generators (Python), tasklets (Stackless Python), backtracking (Prolog) Continuations used to implement cool features like the ones above Coroutines Exceptions Also used in compilers as an intermediate representation no stack needed (e.g. Stackless Python)

6 7 Use case: HTTP HTTP is a stateless protocol Server does not retain information across requests E.g.: Online booking with several forms needs to track progress from page to page typical solutions: cookies, server side sessions, URL-encoding We know better... : Be smart, use continuations! Running example in Chapters of PL book Perl Continuity mimicks stateful HTTP using continuations

7 8 Continuations in Haskell Some languages feature first-class continuations In languages supporting tail-call optimizations and closures, they can be implemented easily Characteristics: a function is in CPS style if it never returns its result is passed to a continuation syntactically: all arguments are λ-expressions or identifiers future of computation as argument continuation maps intermediate result to final result

8 9 Identity direct-style CPS identity id x = x id x k = k x type α α α (α β) β Examples of k: identity or print in direct-style

9 10 Square direct-style CPS f x y k = sq x $ x 2 y 2 f x y = sq x - sq y \xx sq y $ \yy k $ (-) xx yy type Int Int Int Int Int (Int α) α Assume that sq x k is in CPS style

10 Factorial direct-style CPS fac 0 k = k 1 fac 0 = 1 factorial fac n k = fac (n-1) $ fac n = n * fac (n-1) \res k $ (*) n res type Int Int Int (Int α) α

11 Double Transform double f x = f $ f x to CPS, where f is in CPS too! What s the type of double in double f x k? x : α f is in CPS: α (α β) β k continuation: α β double : (α (α β) β) α (α β) β double f x k = f x $ \fx f fx $ \ffx k ffx

12 Regular Expressions Regular expression r over an alphabet Σ defined by r empty language ε empty word a letter a Σ r 1 r 2 concatenation r 1 + r 2 alternation/choice r Kleene star Semantics of regular expression: L(r) Σ String s Σ matches r iff s L(r)

13 15 Regular Expression Matcher Solution Function acc r s k of type Regexp Char String (String Bool) Bool k is a continuation such that for every s Σ, regular expression r and every total function k of type String Bool: if there exists s 1, s 2 Σ such that s = s 1 s 2, s 1 L(r) and k(s 2 ) evaluates to true, then acc r s k evaluates to true if for all s 1, s 2 Σ such that s = s 1 s 2, s 1 L(r) we have that k(s 2 ) evaluates to false, then acc r s k evaluates to false accept r s = acc r s null But: non-termination if ε occurs under a Kleene star

14 16 Fix Program correct for regular expressions in standard form Standard form: ε does not occur under a Kleene star Good news: each regular expression can be transformed into standard form! r = r + δ(r) δ(r) = ε if ε L(r) and otherwise r generates L(r) \ {ε}

15 17 More Fun Proof-directed debugging by Bob Harper available from rwh/papers/regexp/jfp.ps Download this for tutorial! Santa problem: Continuations and Stackless Python: Perl continuity: imagnie HTTP was stateful, see PL book chapters Google tech talk on Parametric Polymorphism, document id on video.google Liquid types director s cut: types full aug09.pptx

16 18 Summary First-class continuations make computations available to explicit manipulation Powerful programming mechanism available in many languages, easily implementable in functional languages At the heart of compilers and behind features like exceptions, fibers, generators Use cases: HTTP and regular expression pattern matcher Next time Equational reasoning (deriving correct programs) Mini test type inference + CPS