Haskell parser to AST data type, assignment

Question

I have been searching around the interwebs for a couple of days, trying to get an answer to my questions and i'm finally admitting defeat.
I have been given a grammar:

Dig ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
Int ::= Dig | Dig Int
Var ::= a | b | ... z | A | B | C | ... | Z
Expr ::= Int | - Expr | + Expr Expr | * Expr Expr | Var | let Var = Expr in Expr

And i have been told to parse, evaluate and print expressions using this grammar
where the operators * + - has their normal meaning
The specific task is to write a function parse :: String -> AST

that takes a string as input and returns an abstract syntax tree when the input is in the correct format (which i can asume it is).

I am told that i might need a suitable data type and that data type might need to derive from some other classes.

Following an example output
data AST = Leaf Int | Sum AST AST | Min AST | ...

Further more, i should consider writing a function
tokens::String -> [String]
to split the input string into a list of tokens
Parsing should be accomplished with
ast::[String] -> (AST,[String])
where the input is a list of tokens and it outputs an AST, and to parse sub-expressions i should simply use the ast function recursively.

I should also make a printExpr method to print the result so that
printE: AST -> String
printE(parse "* 5 5") yields either "5*5" or "(5*5)"
and also a function to evaluate the expression
evali :: AST -> Int

I would just like to be pointed in the right direction of where i might start. I have little knowledge of Haskell and FP in general and trying to solve this task i made some string handling function out of Java which made me realize that i'm way off track.
So a little pointer in the right direction, and maybe an explantion to 'how' the AST should look like
Third day in a row and still no running code, i really appreciate any attempt to help me find a solution! Thanks in advance!
Edit

I might have been unclear: I'm wondering how i should go about from having read and tokenized an input string to making an AST.

Answer 1

Parsing tokens into an Abstract Syntax Tree

OK, let's take your grammar

Dig ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
Int ::= Dig | Dig Int
Var ::= a | b | ... z | A | B | C | ... | Z
Expr ::= Int | - Expr | + Expr Expr | * Expr Expr | Var | let Var = Expr in Expr

This is a nice easy grammar, because you can tell from the first token what sort of epression it will be. (If there was something more complicated, like + coming between numbers, or - being used for subtraction as well as negation, you'd need the list-of-successes trick, explained in Functional Parsers.)

Let's have some sample raw input:

rawinput = "- 6 + 45 let x = - 5 in * x x"

Which I understand from the grammar represents "(- 6 (+ 45 (let x=-5 in (* x x))))", and I'll assume you tokenised it as

tokenised_input' = ["-","6","+","4","5","let","x","=","-","5","in","*","x","x"]

which fits the grammar, but you might well have got

tokenised_input = ["-","6","+","45","let","x","=","-","5","in","*","x","x"]

which fits your sample AST better. I think it's good practice to name your AST after bits of your grammar, so I'm going to go ahead and replace

data AST = Leaf Int | Sum AST AST | Min AST | ...

with

data Expr = E_Int Int | E_Neg Expr | E_Sum Expr Expr | E_Prod Expr Expr | E_Var Char 
                      | E_Let {letvar::Char,letequal:: Expr,letin::Expr}
 deriving Show

I've named the bits of an E_Let to make it clearer what they represent.

Writing a parsing function

You could use isDigit by adding import Data.Char (isDigit) to help out:

expr :: [String] -> (Expr,[String])
expr [] = error "unexpected end of input"
expr (s:ss) | all isDigit s = (E_Int (read s),ss)
             | s == "-" = let (e,ss') = expr ss in (E_Neg e,ss') 
             | s == "+" = (E_Sum e e',ss'') where
                          (e,ss') = expr ss
                          (e',ss'') = expr ss'
            -- more cases

Yikes! Too many let clauses obscuring the meaning, and we'll be writing the same code for E_Prod and very much worse for E_Let. Let's get this sorted out!

The standard way of dealing with this is to write some combinators; instead of tiresomely threading the input [String]s through our definition, define ways to mess with the output of parsers (map) and combine multiple parsers into one (lift).

Clean up the code 1: map

First we should define pmap, our own equivalent of the map function so we can do pmap E_Neg (expr1 ss) instead of let (e,ss') = expr1 ss in (E_Neg e,ss')

pmap :: (a -> b) -> ([String] -> (a,[String])) -> ([String] -> (b,[String]))

nonono, I can't even read that! We need a type synonym:

type Parser a = [String] -> (a,[String])

pmap :: (a -> b) -> Parser a -> Parser b
pmap f p = \ss -> let (a,ss') = p ss 
                  in (f a,ss') 

But really this would be better if I did

data Parser a = Par [String] -> (a,[String])

so I could do

instance Functor Parser where
  fmap f (Par p) = Par (pmap f p)

I'll leave that for you to figure out if you fancy.

Clean up the code 2: combining two parsers

We also need to deal with the situation when we have two parsers to run, and we want to combine their results using a function. This is called lifting the function to parsers.

liftP2 :: (a -> b -> c) -> Parser a -> Parser b -> Parser c
liftP2 f p1 p2 = \ss0 -> let
              (a,ss1) = p1 ss0
              (b,ss2) = p2 ss1
              in (f a b,ss2)

or maybe even three parsers:

liftP3 :: (a -> b -> c -> d) -> Parser a -> Parser b -> Parser c -> Parser d

I'll let you think how to do that. In the let statement you'll need liftP5 to parse the sections of a let statement, lifting a function that ignores the "=" and "in". You could make

equals_ :: Parser ()
equals_ [] = error "equals_: expected = but got end of input"
equals_ ("=":ss) = ((),ss)
equals_ (s:ss) = error $ "equals_: expected = but got "++s

and a couple more to help out with this.

Actually, pmap could also be called liftP1, but map is the traditional name for that sort of thing.

Rewritten with the nice combinators

Now we're ready to clean up expr:

expr :: [String] -> (Expr,[String])
expr [] = error "unexpected end of input"
expr (s:ss) | all isDigit s = (E_Int (read s),ss)
            | s == "-" = pmap   E_Neg expr ss
            | s == "+" = liftP2 E_Sum expr expr ss
            -- more cases

That'd all work fine. Really, it's OK. But liftP5 is going to be a bit long, and feels messy.

Taking the cleanup further - the ultra-nice Applicative way

Applicative Functors is the way to go. Remember I suggested refactoring as

data Parser a = Par [String] -> (a,[String])

so you could make it an instance of Functor? Perhaps you don't want to, because all you've gained is a new name fmap for the perfectly working pmap and you have to deal with all those Par constructors cluttering up your code. Perhaps this will make you reconsider, though; we can import Control.Applicative, then using the data declaration, we can define <*>, which sort-of means then and use <$> instead of pmap, with *> meaning <*>-but-forget-the-result-of-the-left-hand-side so you would write

expr (s:ss) | s == "let" = E_Let <$> var *> equals_ <*> expr <*> in_ *> expr

Which looks a lot like your grammar definition, so it's easy to write code that works first time. This is how I like to write Parsers. In fact, it's how I like to write an awful lot of things. You'd only have to define fmap, <*> and pure, all simple ones, and no long repetative liftP3, liftP4 etc.

Read up about Applicative Functors. They're great.

• Applicative in Learn You a Haskell for Great Good!
• Applicative in Haskell wikibook

Hints for making Parser applicative: pure doesn't change the list. <*> is like liftP2, but the function doesn't come from outside, it comes as the output from p1.

Answer 2

To make a start with Haskell itself, I'd recommend Learn You a Haskell for Great Good!, a very well-written and entertaining guide. Real World Haskell is another oft-recommended starting point.

Edit: A more fundamental introduction to parsing is Functional Parsers. I wanted How to replace a failure by a list of successes by PHilip Wadler. Sadly, it doesn't seem to be available online.

To make a start with parsing in Haskell, I think you should first read monadic parsing in Haskell, then maybe this wikibook example, but also then the parsec guide.

Your grammar

Dig ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
Int ::= Dig | Dig Int
Var ::= a | b | ... z | A | B | C | ... | Z
Expr ::= Int | - Expr | + Expr Expr | * Expr Expr | Var | let Var = Expr in Expr 

suggests a few abstract data types:

data Dig = Dig_0 | Dig_1 | Dig_2 | Dig_3 | Dig_4 | Dig_5 | Dig_6 | Dig_7 | Dig_8 | Dig_9
data Integ = I_Dig Dig | I_DigInt Dig Integ
data Var = Var_a | Var_b | ... Var_z | Var_A | Var_B | Var_C | ... | Var_Z
data Expr = Expr_I Integ 
          | Expr_Neg Expr
          | Expr_Plus Expr Expr
          | Expr_Times Expr Expr Var
          | Expr_Var Var
          | Expr_let Var Expr Expr 

This is inherently a recursively defined syntax tree, no need to make another one. Sorry about the clunky Dig_ and Integ_ stuff - they have to start with an uppercase.

(Personally I'd want to be converting the Integs to Ints straight away, so would have done newtype Integ = Integ Int, and would probably have done newtype Var = Var Char but that might not suit you.)

Once you've done with the basic ones - dig and var, and neg_, plus_, in_ etcI'd go with the Applicative interface to build them up, so for example your parser expr for Expr would be something like

expr =        Expr_I <$> integ 
          <|> Expr_Neg <$> neg_ *> expr
          <|> Expr_Plus <$> plus_ *> expr <*> expr
          <|> Expr_Times <$> times_ *> expr <*> expr
          <|> Expr_Var <$> var
          <|> Expr_let <$> let_ *> var <*> equals_ *> expr <*> in_ *> expr 

So nearly all the time, your Haskell code is clean and closely resembles the grammar you were given.

Answer 3

OK, so it seems you're trying to build lots and lots of stuff, and you're not really sure exactly where it's all going. I would suggest that getting the definition for AST right, and then trying to implement evali would be a good start.

The grammer you've listed is interesting... You seem to want to input * 5 5, but output 5*5, whic is an odd choice. Is that really supposed to be a unary minus, not binary? Simlarly, * Expr Expr Var looks like perhaps you might have meant to type * Expr Expr | Var...

Anyway, making some assumptions about what you meant to say, your AST is going to look something like this:

data AST = Leaf Int | Sum AST AST | Minus AST | Var String | Let String AST AST

Now, let us try to do printE. It takes an AST and gives us a string. By the definition above, the AST has to be one of five possible things. You just need to figure out what to print for each one!

 printE :: AST -> String
 printE (Leaf  x  ) = show x
 printE (Sum   x y) = printE x ++ " + " ++ printE y
 printE (Minus x  ) = "-" ++ printE x
 ...

show turns an Int into a String. ++ joins two strings together. I'll let you work out the rest of the function. (The tricky thing is if you want it to print brackets to correctly show the order of subexpressions... Since your grammer doesn't mention brackets, I guess no.)

Now, how about evali? Well, this is going to be a similar deal. If the AST is a Leaf x, then x is an Int, and you just return that. If you have, say, Minus x, then x isn't an integer, it's an AST, so you need to turn it into an integer with evali. The function looks something like

evali :: AST -> Int
evali (Leaf  x  ) = x
evali (Sum   x y) = (evali x) + (evali y)
evali (Minus x  ) = 0 - (evali x)
...

That works great so far. But wait! It looks like you're supposed to be able to use Let to define new variables, and refer to them later with Var. Well, in that case, you need to store those variables somewhere. And that's going to make the function more complicated.

My recommendation would be to use Data.Map to store a list of variable names and their corresponding values. You'll need to add the variable map to a type signature. You can do it this way:

evali :: AST -> Int
evali ast = evaluate Data.Map.empty ast

evaluate :: Map String Int -> AST -> Int
evaluate m ast =
  case ast of
    ...same as before...
    Let var ast1 ast2 -> evaluate (Data.Map.insert var (evaluate m ast1)) ast2
    Var var           -> m ! var

So evali now just calls evaluate with an empty variable map. When evaluate sees Let, it adds the variable to the map. And when it sees Var, it looks up the name in the map.

As for parsing a string into an AST in the first place, that is an entire other answer again...