MP3 CPS Transform Solution

Logistics




Changelog




1.2




New tests, clarifications and restructuring based on feedback






1.1




Updated deadline; fixed parseExp and parseDecl examples; gave type for cpsDecl.



Objectives




Please read through this entire README before you begin, as some of the early sections may refer to content in later sections. View the PDF version for the most accurate rendering of the notation.




One of the main objectives of this course is to make you comfortable translating code representations from one form to another. In MP2, you wrote an interpreter to evaluate expressions. In this MP you will translate a very small subset of functional-style code from direct style into continuation passing style (CPS). Functional-language compilers can use CPS to express control-flow so that functional language can be compiled to an imperative target.




This will ensure that you understand CPS, and also help you be able to take a mathematical description of a program transformation and implement it in code. You will also practice using a method of propagating state by using a technique known as plumbing.




Goals




Be able to manually translate a chunk of simple functional code into CPS.



Write some functions which can automatically translate simple functional code into CPS.















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Getting Started




Relevant Files




In the directory app you’ll find Main.hs with all the relevant code. In this file you will find all of the data definitions, a simple parser, the REPL, and stubbed-out functions for you to finish.




This README document contains LATEX formatting. To view it properly, look at the README.pdf file you were given, instead of the README.md file that is

shown by default on your GitLab repo page.




Running Code




To run your code, start GHCi with stack ghci (make sure to load the Main module if stack ghci doesn’t automatically). From here, you can test individual functions, or you can run the REPL by calling main. Note that the initial $ and




are prompts.



$ stack ghci




... More Output ...




Prelude :l Main




Ok, modules loaded: Main.




*Main main




To run the REPL directly, build the executable with stack build and run it with stack exec main.




Testing Your Code




As in MP2, you will be able to run the test-suite with stack test:




$ stack test




It will tell you which test-suites you pass, fail, and have exceptions on. To see




an individual test-suite (so you can run the tests yourself by hand to see where the failure happens), look in the file test/Spec.hs.




You can run individual test-sets by running stack ghci --test and choosing to load the Spec module when prompted. Then you can run the tests (specified in test/Tests.hs) just by using the name of the test:




*Spec Main Spec Tests tests_factk




[True,True,True,True,True,True]




Look in the file test/Tests.hs to see which tests were run.










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Given Code




The Types




We have two types for this program. The first is called Stmt, and it corresponds




roughly to a Haskell function declaration. It has one constructor Decl ::




String - [String] - Exp - Stmt. The first argument is the name of the




function, the second is a list of parameter names, and the last is the function body (which is an Exp).




We have also provided a Show instance for the Stmt type, which means that the function show :: Read a = a - String is defined for our type. This




enables pretty-printing so that we can get useful output.




data Stmt = Decl String [String] Exp




deriving (Eq)




instance Show Stmt where




show (Decl f params exp) = f ++ " " ++ intercalate " " params ++ " = " ++ (show exp)




The second type we have is called Exp, which represents expressions. There are six constructors that are encapsulated within Exp. We have IfExp, with the

usual three arguments; AppExp for function application (note: functions are only applied to one of their arguments at a time); IntExp and VarExp for integers and variables; an OpExp that takes an operator (as a string) with two operands; and finally a LamExp to represent anonymous functions.




We’ve also provided a Show instance for the Exp type so that it can be pretty-printed as well.




data Exp = IntExp Integer




VarExp String



LamExp String Exp



IfExp Exp Exp Exp



OpExp String Exp Exp



AppExp Exp Exp deriving (Eq)



instance
Show Exp where




show
(VarExp s)
= s


show
(IntExp i)
= show i


show
(LamExp x e)
= "(\\" ++ x ++ " - " ++ (show e) ++ ")"




show (IfExp e1
e2
e3)
= "(if
" ++
show
e1 ++ "
then " ++ show e2








++ "
else " ++
show e3
++ ")"


show (OpExp op e1
e2)
= "(" ++ show
e1
++""
++ op ++ " " ++ show e2 ++ ")"


show (AppExp f e)


= show f ++ "
" ++ show
e



You may also be interested in seeing how a string is parsed and interpreted







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as something of type Exp. To see how, use the following function ctorParse




String - String. This function will parse the expression normally, but instead of using the default Show instance to display it, it will explicitely show how the Exp is built from data-constructors.



ctorShow :: Exp - String


ctorShow (VarExp s)


= "VarExp " ++ show s
ctorShow (IntExp i)


= "IntExp " ++ show i
ctorShow (LamExp x e)


= "LamExp " ++ show x ++ " (" ++ ctorShow e ++ ")"
ctorShow (IfExp e1 e2
e3)
= "IfExp (" ++ ctorShow e1 ++ ") ("
ctorShow e2 ++ ") ("



ctorShow e3 ++ ")"



ctorShow (OpExp op e1 e2) = "OpExp " ++ show op ++ " ("




ctorShow e1 ++ ") ("



ctorShow e2 ++ ")"



ctorShow (AppExp f e) = "AppExp (" ++ ctorShow f ++ ") (" ++ ctorShow e ++ ")"




fromParse :: Either ParseError Exp - Exp




fromParse (Right exp) = exp




fromParse (Left err) = error $ show err




ctorParse :: String - String




ctorParse = ctorShow . fromParse . parseExp




The Parser




We’ve provided you with a parser again this time. It will live in app/Main.hs with the rest of the code.




First, we have a type synonym to make reading the types of parsers easier.




-- Pretty parser type




type Parser = ParsecT String () Identity




Lexicals




Lexical parsers are just small parsers which look for very simple structure on




the input stream. You’ll notice that the return types of these parsers are all simple types (like Parser String, or Parser Integer). This means that they




don’t have any meaning for our programming language yet, we’ll be using them as building blocks for more complex parsers.




symbol :: String - Parser String




symbol s = do string s




spaces




return s







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int :: Parser Integer




int = do digits <- many1 digit <? "an integer"




spaces




return (read digits :: Integer)




var :: Parser String




var = let keywords = ["if", "then", "else"]




in try $ do v1 <- letter <? "an identifier"




vs <- many (letter <| digit) <? "an identifier"




spaces




let v = v1:vs




if (any (== v) keywords)




then fail "keyword"




else return v




oper :: Parser String




oper = do op <- many1 (oneOf "+-*/<=") <? "an operator"




spaces




return op




parens :: Parser a - Parser a




parens p = do symbol "("




<- p



symbol ")" return pp




Expressions




Now we can write some parsers that are actually relevant to our programming language. You’re not expected to fully understand these parsers yet, or to be able to reproduce them. For your next MP though, you will have to write a simple parser, so it might be worth it to read through and see if it makes sense.




The majority of the complexity in our grammar is with expressions, which is why the parsers for them are more complicated than for lexicals or statements. This is apparent when we look at how many data-constructors we have for Exp, we need to make sure our parser can handle each of them.




intExp :: Parser Exp




intExp = do i <- int




return $ IntExp i




varExp :: Parser Exp




varExp = do v <- var




return $ VarExp v







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opExp :: String - Parser (Exp - Exp - Exp)




opExp str = do symbol str




return (OpExp str)




mulOp :: Parser (Exp - Exp - Exp)




mulOp = opExp "*" <| opExp "/"




addOp :: Parser (Exp - Exp - Exp)




addOp = opExp "+" <| opExp "-"




compOp :: Parser (Exp - Exp - Exp)




compOp = try (opExp "<=") <| try (opExp "=")




<| opExp "<" <| opExp ""




<| opExp "/=" <| opExp "=="




ifExp :: Parser Exp




ifExp = do try $ symbol "if"




e1 <- expr




symbol "then"




e2 <- expr




symbol "else"




e3 <- expr




return $ IfExp e1 e2 e3




lamExp :: Parser Exp




lamExp = do try $
symbol "\\"


param
<- var


symbol "-"


body <- expr


return $ LamExp param body
atom :: Parser Exp


atom =
intExp


<|
ifExp


<|
lamExp


<|
varExp


<| parens
expr
expr :: Parser Exp


expr = let arith
= term `chainl1` addOp


term
= factor `chainl1` mulOp


factor
= app


app
= do f <- many1 atom
in


return $ foldl1 AppExp f
arith `chainl1` compOp



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We’ve also provided the function parseExp :: String - Either ParseError




Exp which can be used to see what the parser returns for a specific input string.




Combined with the function ctorParse that was introduced above, this is useful for understanding how our abstract syntax tree (AST) data structures are being used to store programs written in our language.




parseExp :: String - Either ParseError Exp




parseExp str = parse expr "stdin" str




*Main parseExp "x + 1"




Right (x + 1)




*Main parseExp "x + 1 asdf*" -- This parse will fail.




Left "stdin" (line 1, column 12):




unexpected end of input




expecting white space, an integer, "if", "\\", an identifier or "(" *Main putStrLn $ ctorParse "x + 1" -- Show the AST structure. OpExp "+" (VarExp "x") (IntExp 1)







Declarations




Once we can parse expressions, declarations are comparatively simple.




decl :: Parser Stmt




decl = do f <- var




params <- many1 var




symbol "="




body <- expr




return $ Decl f params body




Once again, we have provided the function parseDecl :: String - Either ParseError Stmt which can be used to investigate the connection between the

grammar of our language and abstract syntax tree of a particular program.

(We have not provided an equivalent of ctorParse for Stmt, but you could try




making your own as an exercise.)




parseDecl :: String - Either ParseError Stmt parseDecl str = parse decl "stdin" str




*Main parseDecl "f x = x + 1"




Right f x = (x + 1)




*Main parseDecl "f x = x + 1 asdf*" -- This parse will fail.




Left "stdin" (line 1, column 18):




unexpected end of input




expecting white space, an integer, "if", "\\", an identifier or "("



















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The REPL




The REPL can be used to enter CPS declarations and see the resulting trans-lations. It will prompt you for string, pass it through the parseDecl parser, check for success, and if so pass it through cpsDecl. Then the result will be pretty-printed using the show :: Show a = a - String function.




prompt :: String - IO ()




prompt str = hPutStr stdout str hFlush stdout




printLn :: String - IO ()




printLn str = hPutStrLn stdout str hFlush stdout




repl :: IO ()




repl = do input <- prompt " " getLine




case input of




"quit" - return ()




- do case parseDecl input of



Left err - do printLn "Parse error!"




printLn $ show err




Right decl - printLn . show $ cpsDecl decl




repl







main :: IO ()




main = do putStrLn "Welcome to the CPS Transformer!"




repl




putStrLn "GoodBye!"







Problems




Manual Translation




First, you’ll manually translate a few Haskell functions into CPS. This can help you gain an intuition about how the automatic translator will work.




factk :: Integer - (Integer - t) - t




Write the factorial function in continuation passing style. The first argument is the number that it should compute the factorial of. The second argument is the continuation to apply to the result.




Note that you will not receive credit if you calculate the factorial separately using a normal factorial function then put it through the continuation. You




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should instead recursively cal factk with an updated continuation which contains the computation to make at this step in the recursion as well as the original continuation.




*Main Lib factk 10 id




3628800







evenoddk :: [Integer] - (Integer - t) - (Integer - t) - t




Write the function evenoddk in continuation passing style.




The evenoddk function takes a list and two continuations. The first continuation, if run, will receive the sum of all the even elements of the list. The second continuation, if run, will receive the sum of all the odd elements of the list.




No additions should be performed until a continuation is called. This means that you need to build up the addition to make in each continuation, not do the addition as you process the list.




The function decides what to do when it gets to the last element of the list. If it is even, it calls the even continuation, otherwise it calls the odd continuation.




You can assume the input list will have at least one entry in it.




*Main Lib evenoddk [2,4,6,1] id id




1




*Main Lib evenoddk [2,4,6,1,4] id id




16




*Main Lib evenoddk [2,4,6,1,9] id id




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Automated Translation




Here, you’ll define two functions do all the translation. We will call cpsDecl on a declaration (Stmt). This function will return a new declaration with an added argument k, and a body which has been CPS transformed to use that as its continuation.




We’ll also have a function cpsExp which is used to translate an Exp into CPS. It takes an extra argument, which is an integer, and we return an integer in addition to the transformed expression. These integers are how we are going to get fresh variables each time we want to generate a new continuation. They act as a sort of accumulator, receiving the number for the next fresh variable and returning a potentially updated number after the transformation is performed.




In particular, we have given you a utility function called gensym (for “generate




symbol”) that generates fresh variables each time it is called using that particular integer. Whenever we need a fresh variable, we call gensym with our old integer.




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gensym :: Integer - (String, Integer)




gensym i = ("v" ++ show i, i + 1)




*Main Lib gensym 20




("v20",21)




Notice that the return type for both gensym and cpsExp is a tuple of an expected return type and an integer; this is how we remember what numbers have already been used (and what comes next). This is a common pattern in languages that do not have mutation (or even languages that do but we chose not to use it), and is often called “plumbing.”




The CPS transform [[D]] represents a program transform taking declaration D from direct style to continuation passing style. The CPS transform [[E]]k repre-sents a program transform taking expression E from direct style to continuation passing style with k as its continuation. Remember that the continuation is a function to apply to the result of the current expression. In this example, we will call the continuation k on the value that expression E evaluates to.




We will also distinguish between simple expressions (no available function calls) and normal expressions. (We will go into more detail later about how we will distinguish simple from normal expressions.) Operators will be left in direct style, and we will not be converting λ-expressions in this MP (though your code will certainly emit them!).1




To indicate that an expression is simple, we will put an overbar on it, like this: e. Some expressions, like variables and integers, are obviously simple so we will omit the overbar in such a case. As an example, e1 + e2 indicates that e1 is simple, but that e2 is not. Thus, e2 would need to be transformed.







Define isSimple




The isSimple function takes an expression and determines if it is simple or not. A simple expression, in our context, is one that does not have an available function call. A function call is available if it’s possible for the current expression to execute it. In the language subset you have been given, a function call is always available unless it is within the body of an unapplied λ-expression. We




aren’t transforming λ-expressions in this MP, so you can ignore implementing isSimple over LamExps.




*Main Lib isSimple (AppExp (VarExp "f") (IntExp 10))




False




*Main Lib isSimple (OpExp "+" (IntExp 10) (VarExp "v"))







Most automatic CPS transforms do not make this distinction: there are, in fact, many different CPS transforms, and researchers have spent a lot of time coming up with transforms that have different properties. This means that if you find a paper describing a CPS transform, odds are excellent it will not look exactly like this one. You should still find it recognizable though!






10















True




*Main Lib isSimple (OpExp "+" (IntExp 10) (AppExp (VarExp "f") (VarExp "v")))




False




Define cpsExp - Overview




You’ll need to define cpsExp over all of the data-constructors that make up the Exp type. The first argument to cpsExp is the expression you are transforming,

and the second is the current continuation that has been built up. Here is the type signature:




cpsExp :: Exp - Exp - Integer - (Exp, Integer)




Remember that we have provided you with a parser and the function parseExp,




which you can use to type in expressions for testing. Instead of having to write the entire Haskell AST of an expression, you can use parseExp.




Define cpsExp for Integer and Variable Expressions




[[i]]k = k i

[[v]]k = k v




In Haskell, this would (almost) be implemented as:




almostCpsExp :: Exp - Exp - Exp




almostCpsExp (IntExp i) k = AppExp k (IntExp i)




almostCpsExp (VarExp v) k = AppExp k (VarExp v)




But, remember that we have to add in the extra “plumbing” which keeps track of how many fresh variables we’ve generated, which accounts for the extra Integer in both the arguments and return of cpsExp.




Define cpsExp for Application Expressions




[[f


]]k
= f e k
e
[[f
e]]k
= [[e]](λv.f v k) where v is fresh.



Note there are two rules: one for when the argument is simple, and one for when the argument needs a conversion. We are going to assume that the argument is run first, the result of which is passed into the function.




Remember that function application is left-associative. That is, assuming that we use currying, f e k = ((f e) k). Also recall that in our terminology, we “apply” functions to arguments. We do not “apply” arguments to functions.










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Notice the “where v is fresh” part. Variables don’t grow moldy, but if you give every continuation function a parameter named v, it won’t work. We could have nested continuations which lead to unwanted α-capture. Therefore, we need to generate new names whenever we make a new λ-expression. The safest strategy is to make sure that no name is ever reused, regardless of which branch of logic is eventually taken.




This should be done using the gensym :: Integer - (String, Integer) function explained above. Remember that cpsExp needs to return the most up-to-date fresh variable number so that further calls to cpsExp don’t use the




same variables again.




Define cpsExp for Operator Expressions




These are the most complex, since there are four possible cases.




For ⊗ a binary operator:




[[




⊗


]]k = k (e1 ⊗ e2)


e1
e2


[[e1
⊗


]]k = [[e1
]](λv.k(v⊗e2)
where v is fresh.
e2
[[


⊗ e2]]k = [[e2
]](λv.k(e1⊗v)
where v is fresh.
e1
[[e1
⊗ e2]]k = [[e1
]](λv1.[[e2]](λv2.k(v1⊗v2))) where v1 and v2 are fresh.



Notice that we are careful to preserve the order of the arguments to the operator




the order we convert expressions to is very important (that is, e1 and e2 cannot be swapped on the right-hand side of any of the above), because this preserves the order of evaluation. In real life it doesn’t matter the order we convert them in (that is, whether we choose to convert e1 or e2 first), as long as their results are passed in the correct order, but to keep from annoying the graders2 please evaluate things from left to right.



Define cpsExp for If Expressions







[[if


then e2
else e3]]k
= if e1 then [[e2]]k else [[e3]]k
e1
[[if e1
then e2
else e3]]k
= [[e1]](λv.if v then [[e2]]k else [[e3]]k) where v is fresh.



For an if expression with a not-simple guard, we need to transform the guard and give it a continuation that selects the proper (transformed) branch for us. If the guard is simple, we can leave it alone and transform the branches using the original continuation.




Notice how the order of evaluation becomes explicit when we make this transform; the condition of the if expression is evaluated first, and only then is one of the




Annoying a grader is bad luck.



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branches picked for evaluation. This is part of the power of CPS, you can choose exactly how expressions in your language are simplified down.




Define cpsDecl




Declarations are much simpler than expressions because we don’t need to maintain the state of a fresh-variable counter when translating declarations. This is because each declaration has its own local variable scope anyway.




[[f x1 · · · xn = e]] = (f x1 · · · xn k = [[e]]k)




This corresponds to the cpsDecl function. It adds an extra parameter k to the parameter list, and then translates the body of the function. (The parentheses




on the right-hand side are added for clarity - the equality on the left tranforms to the whole equality on the right.) Here is the type signature for cpsDecl:




cpsDecl :: Stmt - Stmt




Monads (optional)




This part is highly optional.




Perhaps you are thinking “couldn’t we use a monad to get rid of this plumbing?” If so, you are correct. There is a monad called the State monad that can do this for us. (In fact, there is a GenSym monad as well!)




You are not required to use monads for this MP. But if you would like to use




them, you may, as long as you do not change the type of cpsDecl or cpsExp. Define your own function cpsExpM, the monadic version of cpsExp, and you can define cpsExp in terms of cpsExpM instead. The testing interface to cpsExp




must remain unchanged.























































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