How To Write an Interpreter for a Lambda-Calculus-Based Language

1. Value type

You’ll need a type to represent the values your language works on, a sum of all the types of values you want to work with. Since the language is based on the lambda calculus, it will need to include a constructor for functions. For example:

data Value =
    VBool Bool |
    VText Text |
    VInt Int |
    VList [Value] |
    VFunction (Value -> Value) |
    VError String

Note that we’re handling errors as a kind of value (VError), but there are other approaches to error-handling.

2. Apply function

This is a function takes a “function” value and an “argument” value and applies them. It’ll probably look something like this:

vapply :: Value -> Value -> Value
vapply (VFunction f) a = f a
vapply (VError err) _ = VError err
vapply _ _ = VError "not a function" -- or a more informative error message

Note how we pass through errors that are being used as functions. if we want to, we could also pass through errors that are being used as arguments, making all functions “strict”.

3. The expression type

You’ll need a type for expressions. An expressions has some set of free (unbound) variables, which can then be bound in composition with other expressions. For example:

f a 3 has free variables f a

let p=q in p has free variables q

(\a -> a) 1 has no free variables

Expressions with no free variables are “closed”, those with one or more are “open”.

Here’s the magic type for expressions with lexical scope:

data Expr a = ClosedExpr a | OpenExpr Name (Expr (Value -> a))

There are lots of variations and generalisations of this type, but this is the basic form. (Name is of course the type of names of variables, probably a synonym for String or Text.) Here’s how it represents expressions, all of type Expr Value:

text	value
`4`	`ClosedExpr $ VInt 4`
`a`	`OpenExpr "a" $ ClosedExpr $ \a -> a`
`f 1`	`OpenExpr "f" $ ClosedExpr $ \f -> vapply f (VInt 1)`
`f a`	`OpenExpr "f" $ OpenExpr "a" $ ClosedExpr $ \f a -> vapply f a` — be careful about argument order

4. instance Applicative Expr

You’ll need to create Functor and Applicative instances for your expression type, which I’ll leave for you to do. It’s easy to get the argument ordering wrong, but we’ll catch this with tests. The (<*>) function gives you the “application” operation of the lambda calculus, while pure creates constant expressions.

5. var, abstract and let

You’ll need to create functions for the other lambda calculus operations:

exprVar :: Name -> Expr Value
exprAbstract :: Name -> Expr a -> Expr (Value -> a)
exprLet :: Name -> Expr Value -> Expr a -> Expr a

It’s easy to write exprLet using exprAbstract, you simply make this transformation:

let var = bind in val

⇒

(\var -> val) bind

Again, if you implement these wrong your tests will catch them.

6. Recursive let-binding

Of course, this let-binding operation doesn’t allow recursion. The definition cannot refer to the name being defined, nor can you have a set of definitions that refer to each other cyclically. This might be what you want, but if it isn’t, no worries, Haskell is a lazy language and it’s straightforward to implement recursive let-binding (letrec) in terms of sequential let-binding (letseq).

To show how, let’s first see how it works in the case of a single binding. It’s simply this transformation:

letrec var = bind in val

⇒

letseq var = fix (\var -> bind) in val

Now for a set of bindings, the idea is to gather them together into a single binding, and perform the same transformation:

letrec var0=bind0; var1=bind1; var2=bind2 in val

⇒

letseq
vars = fix (\vars -> let var0=extract0 vars; var1=extract1 vars; var2=extract2 vars in join bind0 bind1 bind2);
var0=extract0 vars;
var1=extract1 vars;
var2=extract2 vars
in val

Of course, you’ll need to determine a suitable type for vars, with corresponding join and extractN functions, paying attention to strictness and irrefutability in matching.

7. Evaluation

Evaluation is turning an expression into a resulting value. This is pretty easy: if an expression is closed, you have a result; if it’s open, then variables have been left unbound:

exprEvaluate :: Monad m => Expr a -> m a
exprEvaluate (ClosedExpr a) = return a
exprEvaluate (OpenExpr name _) = fail $ "unbound variable: " ++ name -- or list all of them instead

8. Parsing, generally

So now you’ll need to parse text into expressions. The parsec library is well-suited for this; the Parser monad basically does what you want. The main caveat is how it handles failure in choice:

p <|> q: if p fails and consumed no input, then q (otherwise propagate p ‘s failure)

try p <|> q: if p fails then q

The key is to pay attention to the first character to be parsed in each branch of a choice. If they’re all different, you don’t need to use try .

9. Parsing whitespace

First thing is to create a parser for whitespace, including comments, which we’ll call readWS . After that, you can use the convention that all parsing functions munch trailing (but not leading) whitespace. I generally find these functions useful:

readCharAndWS :: Char -> Parser ()
readCharAndWS c = do
    _ <- char c
    readWS

readStringAndWS :: String -> Parser ()
readStringAndWS s = do
    _ <- string s
    readWS

Of course, your top-level parser will also have to munch leading whitespace.

parse :: Parser (Expr Value)
parse = do
    readWS -- munch leading whitespace
    readExpression

10. Parsing identifiers

It’s up to you of course, but names of variables and other identifiers typically start with a letter, followed by zero-or-more alphanumeric characters. You might want to also include underscores and some other characters. Use the character-test functions in Data.Char and you’ll get Unicode support for free.

If you have keywords in your language then you must specifically test and exclude them from identifiers, otherwise you’re going to have a bad time.

Don’t forget to add a trailing readWS !

11. Parsing expressions

You’ll want to pay careful attention to ambiguity and precedence, with functions for parsing “loose” and “tight” expressions. For numerical and other infix expressions, you can use the notion of “terms” (which contribute to a sum) and “factors” (which contribute to a product). For example, the sum a + b * c is compose of two terms, the latter of which is a product of two factors. For example:

readExpression :: Parser (Expr Value)
readExpression = ... readTerm ...

readTerm :: Parser (Expr Value)
readTerm = ... readFactor ...

readFactor :: Parser (Expr Value)
readFactor = ... parenthetical readExpression ...

12. Predefined functions

You’ll probably want to supply a set of bindings to useful functions, that you’ll need to convert to your Value type. The easiest way to do this is to create FromValue and ToValue classes:

class FromValue t where
    fromValue :: Value -> Either String t
class ToValue t where
    toValue :: t -> Value

-- Int example instance
instance FromValue Int where
    fromValue (VInt v) = Right v
    fromValue (VError err) = Left err
    fromValue _ = Left "expected int"
instance ToValue Int where
    toValue = VInt

Then you can create a system of instances that allows easy conversions of Haskell functions.

instance FromValue Value
instance ToValue Value
instance (ToValue a,FromValue b) => FromValue (a -> b)
instance (FromValue a,ToValue b) => ToValue (a -> b)
instance FromValue Int
instance ToValue Int
instance FromValue Text
instance ToValue Text

… etc.

You can then let-bind your library of values to expressions created by the user.

13. Putting it all together

Now you have all the pieces, the pipeline looks like this:

Parse the user’s text to get an expression (of type Expr Value).
Let-bind the library of predefined values.
Evaluate the result to get a value (of type Value).

And that’s it!

14. Tests

You’ll want hundreds of these. I use tasty-hunit as a basic test framework. The good news is that it’s very easy to write lots, as the basic form is just a language string and an expected result (fail, or pass with this value). The only awkwardness is that you probably don’t have an instance Eq Value, because of its function constructor, so you’ll need to create a value-comparison function, or even an orphan Eq instance just for testing.

Be sure to test

whitespace,
comments,
boolean literals,
text literals,
numeric literals,
list construction,
parsing ambiguity,
function abstraction,
function application,
let-binding,
recursive let-binding,
scoping,
predefined functions,
error propagation,
anything else you can think of…

And if you’re really serious, you can use QuickCheck to generate language strings and expected results.

— Ashley Yakeley

With-Resource Puzzle

Haskell Maxims and Arrows

There is No Haskell Topology

4 Replies

Hey, it’s Haskell Hwednesday! For those learning Haskell who are already educated in mathematics, or vice versa, the experience can be a bit of a rollercoaster.

We first discover the category of Haskell types and functions, predictably called Hask. This is a very nice sort of category: it has a terminal object (the unit type), products (pair tuple types) and exponentiation (function types), indeed a well-known sort of category known as Cartesian closed. For good measure, it also has sums (Either types) and an initial object (any empty type). Fabulous! A programming language with the beauty and rigour of mathematics.

Then we discover that almost none of this is true. Hask is indeed a category, but it does not have products or sums, and the unit type is not a terminal object, and there is no initial object. It turns out we actually had a different category in mind, a kind of IdealHask. In our IdealHask, typically, all functions terminate. In the actual Hask, every object is a Haskell type, and every Haskell type includes “bottom values” representing non-termination and other awkward conditions. (There’s actually a faithful functor from IdealHask to Hask that allows us to write our programs.)

Oh dear. These “extra” values have ruined everything. Dan Piponi was right, you shouldn’t have even asked.

But wait! As it happens, simple non-termination can be represented by topologies, which are awesome. We associate a topology with every type, and the definable functions are exactly the continuous ones. For example, the type Bool nominally has two values, True and False. We add non-termination as a third value bottom. The topology we define on this has these open sets: {{}, {True}, {False}, {True, False}, {True, False, bottom}}. The eleven definable functions are the nine strict ones (i.e., where f bottom = bottom):

\x -> if x then True else True
\x -> if x then True else False
\x -> if x then True else bottom
\x -> if x then False else True
\x -> if x then False else False
\x -> if x then False else bottom
\x -> if x then bottom else True
\x -> if x then bottom else False
\x -> if x then bottom else bottom

and the two non-strict ones:

\_ -> True
\_ -> False

These are precisely the continuous functions from our topological space to itself. And it seems perhaps straightforward to generalise this to any type?

Yay topology! We have some of our mathematicalliness back! We think.

Ah, but Haskell types are more complicated than this. There are many different bottom values, that are included in every type. Non-termination is one, but there are also exceptions. Plus, there’s a function called mapException that can inspect exception values and arbitrarily map them to bottom values. Can we create a topology for each Haskell type such that the definable functions are exactly the continuous ones?

Sadly, we cannot. To show this, let us consider different models of Haskell’s unit type (), and what can happen when we attempt to evaluate a term of that type, which we’ll collectively refer to as possibilities. We’ll define the models as types in Haskell, specifying exactly which functions are definable. Then we’ll search exhaustively for topologies that match. You’ll need to install the countable package, which among other things includes an extensional “instance Eq (a -> b)“.

{-# LANGUAGE ScopedTypeVariables, TypeSynonymInstances, FlexibleInstances, OverlappingInstances #-}
module Main where

import Control.Applicative (liftA,liftA2,liftA3)
import Data.List (intercalate)
import Data.Countable
import Data.Searchable

First some set theory:

implies :: Bool -> Bool -> Bool
implies False _ = True
implies True x = x

type Set x = x -> Bool -- is it a member?

empty :: Set x
empty _ = False

full :: Set x
full _ = True

union :: Set x -> Set x -> Set x
union p q a = p a || q a

intersection :: Set x -> Set x -> Set x
intersection p q a = p a && q a

instance (Finite x,Show x) => Show (Set x) where
    show s = "{" ++ (intercalate "," (fmap show (filter s allValues))) ++ "}"

And this is how we do topology:

-- Given a function, find the preimage of a set.
preimage :: (x -> y) -> Set y -> Set x
preimage f s a = s (f a)

type Topology x = Set (Set x) -- set of open sets

isTopology :: (Countable x) => Topology x -> Bool
isTopology top = hasEmpty && hasFull && hasIntersections && hasUnions where
    hasEmpty = top empty
    hasFull = top full
    hasIntersections = forevery (\ (p,q) -> (top p && top q) `implies` top (intersection p q))
    hasUnions        = forevery (\ (p,q) -> (top p && top q) `implies` top (union p q))

-- Given topologies on x and y, find the set of continuous functions
continuous :: (Countable y) => Topology x -> Topology y -> Set (x -> y)
continuous topx topy f = forevery (\set -> implies (topy set) (topx (preimage f set)))

Now we need to model Haskell types in Haskell:

class (Finite x) => Model x where
    -- definable functions in this model
    definable :: Set (x -> x)

The simplest model of the unit type has only one possibility, termination with the unit value, in a language where all functions terminate. We’ll call that model U1, and its sole possibility T1. The only possible function there can be maps T1 to T1, and it is definable:

-- U1 models the unit type in a strict language
-- T1 represents the only possibility, termination with the unit value
data U1 = T1 deriving (Eq, Show)

instance Countable U1 where
    countPrevious = finiteCountPrevious
    countMaybeNext = finiteCountMaybeNext

instance Searchable U1 where
    search = finiteSearch

instance Finite U1 where
    allValues = [T1]
    assemble afb = liftA (\t a -> case a of
        T1 -> t
        ) (afb T1)

instance Model U1 where
    definable f = af (f T1) where
        af T1 = True  -- there is only one possible function, which is definable

Our second model, U2, adds a non-termination possibility N2 (termination with the unit value is T2). A function is definable if either

it ignores its input and behaves the same whatever it is passed
it examines its input and therefore doesn’t terminate when the input doesn’t terminate

-- U2 represents the unit type in a lazy language with no exceptions besides non-termination
-- T2 is termination with the unit value,
-- N2 is non-termination
data U2 = T2 | N2 deriving (Eq, Show)

instance Countable U2 where
    countPrevious = finiteCountPrevious
    countMaybeNext = finiteCountMaybeNext

instance Searchable U2 where
    search = finiteSearch

instance Finite U2 where
    allValues = [T2,N2]
    assemble afb = liftA2 (\t n a -> case a of
        T2 -> t
        N2 -> n
        ) (afb T2) (afb N2)

instance Model U2 where
    definable f = af (f T2) (f N2) where
        af T2 T2 = True  -- f _ = always termination
        af N2 N2 = True  -- f _ = always non-termination
        af T2 N2 = True  -- f x = x
        af N2 T2 = False -- "inversion", not definable

Our third model, U3, adds an “exception” possibility. An exception is a termination, but still considered by Haskell to be “bottom”. The previous rule applies, but in addition, Haskell ensures that if a function examines its input, it must yield a bottom when passed a bottom. However, it’s allowed to map exception to a different bottom using mapException.

-- U3 represents the unit type in a lazy language with an additional "exception" possibility
-- T3 is termination with the unit value,
-- E3 is termination with the exception
-- N3 is non-termination
-- (E3 and N3 are both "bottoms")
data U3 = T3 | E3 | N3 deriving (Eq, Show)

instance Countable U3 where
    countPrevious = finiteCountPrevious
    countMaybeNext = finiteCountMaybeNext

instance Searchable U3 where
    search = finiteSearch

instance Finite U3 where
    allValues = [T3,E3,N3]
    assemble afb = liftA3 (\t e n a -> case a of
        T3 -> t
        E3 -> e
        N3 -> n
        ) (afb T3) (afb E3) (afb N3)

instance Model U3 where
    definable f = af (f T3) (f E3) (f N3) where
        -- functions that ignore the argument are definable
        af t e n | n == e && n == t = True
        -- otherwise, the function must not terminate when passed non-termination:
        af _ _ T3 = False
        af _ _ E3 = False
        -- more generally, the function must return a bottom when passed a bottom: 
        af _ T3 _ = False
        -- all other functions are definable
        af _ _ _ = True

Now we want to search for topologies. Thanks to the countable package, we can do this exhaustively.

-- this topology matches if its continous functions are exactly the definable functions
matching :: (Model x) => Topology x -> Bool
matching top = (continuous top top) == definable

-- all matching topologies
goodtops :: (Model x) => [Topology x]
goodtops = filter (\top -> matching top && isTopology top) allValues

printOnLines :: (Show s) => [s] -> IO ()
printOnLines = mapM_ (putStrLn . show)

main :: IO ()
main = do
    putStrLn "Topologies for U1:"
    printOnLines (goodtops :: [Topology U1])
    putStrLn "Topologies for U2:"
    printOnLines (goodtops :: [Topology U2])
    putStrLn "Topologies for U3:"
    printOnLines (goodtops :: [Topology U3])

And that’s it. This is what we get when we run our program:

Topologies for U1:
{{},{T1}}
Topologies for U2:
{{},{T2},{T2,N2}}
{{},{N2},{T2,N2}}
Topologies for U3:

U1 has the one-point topology. U2 has the Sierpiński topologies. But there is no matching topology for U3.

So what about the actual Haskell unit type? Consider the equivalence relation R with three equivalence classes: the unit value, non-termination, and all exceptions, corresponding to the possibilities of U3. For each “definable” function U3 -> U3, we can define a Haskell function () -> () that corresponds to it under R. For “disallowed” functions, we cannot.

So if there were a topology X on the Haskell unit type such that the continuous functions were the definable ones, then X/R would be a quotient topology on U3 where the continuous functions were the definable ones. But this doesn’t exist, so that doesn’t either.