This is a Homework Problem.
My goal is to Convert a type Expression in the form of " into a list of CPU Instructions. Given the data structures
data Expr = Const Double | Var String | Add Expr Expr | Mult Expr Expr
data Instr = LoadImmediate RegisterNumber -- put value here
RegisterValue -- the value
| Addition RegisterNumber -- put result here
RegisterNumber -- first thing to add
RegisterNumber -- second thing to multiply
| Multiply RegisterNumber -- put result here
RegisterNumber -- first thing to multiply
RegisterNumber -- second thing to multiply
type RegisterNumber = Int
type RegisterValue = Double
Type Expression has four main functions
Const: converts a number of type double to a type expression, letting you use it.
Var: basically converts a string (i.e. "x") into a type expression letting you apply it to constant
Then then Add and Mult commands that let you add and multiply two type expressions
And we can assume that the only variable we will see is "x" and it is already in register 2. The result will arrive in register 1. There is a total of 4 registers.
So Add Const 1 (Mult (Var "x") (Const 2))
in type [Instr] would be
[LoadImmediate 3 1,
LoadImmediate 4 2,
Multiply 1 4 2,
Addition 1 3 1]
EDIT: Sorry, I for got to mention, because this is a beginner course, we only need to consider expressions of the form
(Const a) `Add` ((Var "x") `Mult` Expr)
where 'Expr' is some expression. Or in math form a0+x(a1+x(a2+x...))
I fixed my code up a little bit, now the error I'm getting is "Not in scope: data constructor 'RegNum'"
expr2Code :: Expr -> [Instr]
expr2Code expr = e2C 1 expr
e2C:: Int -> Expr -> Instr
e2C RegNum (Const y) = [LoadImmediate RegNum y]
e2C RegNum (Var "x") = []
e2C RegNum (Add expr1 expr2) = e2C 3 expr1 ++ e2C 4 expr2 ++ [Addition RegNum 3 4]
e2C RegNum (Mult expr1 expr2) = e2C 3 expr1 ++ e2C 4 expr2 ++ [Multiply RegNum 3 4]
Sorry for the long post, any help would be appreciated.
Well I'm assuming you have an infinite number of registers. If not you can experience the joy that is register spilling, but you'd need some more instructions to deal with dynamic memory.
There are 3 straightforward ways to do this
Expressions -> SSA -> Instr
Expressions -> CPS -> Instr
Expressions -> Instr
The first 2 offer much easier opportunities to optimize your use of registers and what not, but involve an intermediate language. Since we're lazy, let's do 3.
import Control.Monad.State
type Gensym = State Int
gensym :: Gensym RegisterNumber
gensym = modify (+1) >> fmap (subtract 1) get
Now that we have a way of uniquely generating registers, we can do the wonderfully inefficient approach.
withRegister :: (RegisterNumber -> [Instr]) -> Gensym (RegisterNumber, [Instr])
withRegister f = gensym >>= \r -> return (r, f r)
compile :: Expr -> Gensym (RegisterNumber, [Instr])
compile (Const d) = withRegister $ \r -> [LoadImmediate r d]
compile (Var "x") = return (2, [])
compile (Add e1 e2) = do
(t1, c1) <- compile e1 -- Recursively compile
(t2, c2) <- compile e2 -- Each subexpression like you were
withRegister $ \r -> Addition t1 t2 r : (c1 ++ c2) -- Make sure we
-- use a unique register
compile (Mult e1 e2) = do
(t1, c1) <- compile e1
(t2, c2) <- compile e2
withRegister $ \r -> Multiply t1 t2 r : (c1 ++ c2)
compileExpr :: Expr -> [Instr]
compileExpr = reverse . snd . flip evalState 3 . compile
This basically recursively compiles each expression, concatting their various chunks of code together. This is similar to what you had, but there are 3 major differences,
I'm using a monad to handle the registers for me.
You have to ensure that you never clobber a register you're going to need, by using a monad I ensure that all the registers I'm using are unique. This is really inefficient,but trivially correct.
When handling Var, I just load whatever's in register 2, since LoadImmediate wants a double and you have no mechanism for actually binding variables in your expressions.
Because you're not dealing with expressions, each chunk of computation has to be stuck in a register explicitly. You can't do x + y * z any more. That's why if you look at the code for Add or Mult, each subexpression is compiled to a fresh register.
Your example generates
[LoadImmediate 4 2.0,Multiply 2 4 5,LoadImmediate 3 1.0,Addition 3 5 6]
Which is correct, it multiplies 4 and x, then adds 3.
e2C _ (Var "x") = LoadImmediate 2 "x"
If x is already in the register 2 you don't need to load it at all. Var "x" would not translate to any load operation. rather, it translates to the operand of 2 in some other operation (addition or multiplication). For instance, (Add (Const 25) (Var "x")) would translate to [LoadImmediate 3 25, Addition 1 3 2].
e2C _ (Mult x y) = Multiply 4 (e2C _ x) (e2C _ y)
This of course does not work, as operands of Multiply are registers, not instructions. You have to translate x and note to which register rx the result goes; then translate y and note to which register ry its result goes; make sure rx != xy; and finally, issue a Multiply rz rx ry.
Now, how to determine rz, and how to make sure rx != ry? One simple strategy is to make sure each result goes to its own register (assuming there's an infinite number of them, which is of course not true for real machine architectures).
The top-level result will go to register 1.
Related
I'm trying to understand Haskell monads and wrote this test program, which compiles and works as expected:
divide :: Int -> Int -> Either String Int
divide _ 0 = Left "Divide by zero error
divide numerator denom = Right (numerator `div` denom)
processNumsMonadically :: Int -> Int -> Either String Int
processNumsMonadically n d = divide n d >>= \q -> return (q+1)
When I try using the word bind instead of the >>= operator in the latter function definition:
processNumsMonadically n d = bind (divide n d) (\q -> return (q+1))
I get the error:
Not in scope: 'bind'
What is the correct way to use the word bind?
This isn't a part of Prelude; it resides in Control.Monad.Extra, a part of monad-extras package.
However, you can call operators in prefix manner (like named functions) easily:
processNumsMonadically n d = (>>=) (divide n d) (\q -> return (q+1))
You could also just use do notation:
processNumsMonadically n d = do
q <- divide n d
return (q+1)
But while we're at it, I'd write using a Functor:
processNumsMonadically n d = fmap (+1) (divide n d)
or Applicative syntax:
processNumsMonadically n d = (+1) <$> divide n d
You could also lift the +1 to avoid the need for return and the lambda.
As a personal style remark, bind used as a word isn't idiomatic, and IMHO you shouldn't use it.
There is GenProg (http://hackage.haskell.org/package/genprog) for example, but that only deals with numerical optimization, in this case finding an equation that describes the data.
But I require for loops, if statements, when statements, Boolean checks etc. I need to be able to generate imperative structures. Any thought on this? My best options so far seem to be husk-scheme where I can run Scheme code as a DSL in Haskell. Surely there must be better ways?
If you're looking for something akin to S-expressions, this is fairly easily modeled in Haskell. Say, for example, we want to represent simple algebraic equations with variables, such as
x + 5 / (y * 2 - z)
This can be represented by a simple AST in Haskell, in particular we can implement it as
data Expr
= Lit Double -- Literal numbers
| Var Char -- Variables have single letter names
| Add Expr Expr -- We can add things together
| Sub Expr Expr -- And subtract them
| Mul Expr Expr -- Why not multiply, too?
| Div Expr Expr -- And divide
deriving (Eq)
This would let us express the equation above as
Add (Var 'x') (Div (Lit 5) (Sub (Mul (Var 'y') (Lit 2)) (Var 'z')))
But this is rather clunky to write and difficult to read. Let's start by working some Show magic so that it gets pretty printed:
instance Show Expr where
showsPrec n (Lit x) = showParen (n > 10) $ showsPrec 11 x
showsPrec n (Var x) = showParen (n > 10) $ showChar x
showsPrec n (Add x y) = showParen (n > 6) $ showsPrec 7 x . showString " + " . showsPrec 7 y
showsPrec n (Sub x y) = showParen (n > 6) $ showsPrec 7 x . showString " - " . showsPrec 7 y
showsPrec n (Mul x y) = showParen (n > 7) $ showsPrec 8 x . showString " * " . showsPrec 8 y
showsPrec n (Div x y) = showParen (n > 7) $ showsPrec 8 x . showString " / " . showsPrec 8 y
If you don't understand everything going on here, that's ok, it's a lot of complication made easy by some built in functions for efficiently building strings with parentheses in them properly and all that fun stuff. It's pretty much copied out of the docs in Text.Show. Now if we print out our expression from above, it'll look like
x + 5.0 / (y * 2.0 - z)
Now for simplifying building these expressions. Since they're pretty much numeric, we can implement Num (except for abs and signum) and Fractional:
instance Num Expr where
fromInteger = Lit . fromInteger
(+) = Add
(-) = Sub
(*) = Mul
abs = undefined
signum = undefined
instance Fractional Expr where
(/) = Div
fromRational = Lit . fromRational
Now we can input out expression from above as
Var 'x' + 5 / (Var 'y' * 2 - Var 'z')
This is at least much easier to visually parse, even if we have to specify the variables manually.
Now that we have pretty input and output, let's focus on evaluating these expressions. Since we have variables in here, we'll need some sort of environment that associates a variable with a value
import Data.Map (Map)
import qualified Data.Map as M
type Env = Map Char Double
And now it's just basic pattern matching (along with a helper function)
import Control.Applicative
binOp :: (Double -> Double -> Double) -> Expr -> Expr -> Env -> Maybe Double
binOp op x y vars = op <$> evalExpr x vars <*> evalExpr y vars
evalExpr :: Expr -> Env -> Maybe Double
evalExpr (Lit x) = const $ Just x
evalExpr (Var x) = M.lookup x
evalExpr (Add x y) = binOp (+) x y
evalExpr (Sub x y) = binOp (-) x y
evalExpr (Mul x y) = binOp (*) x y
evalExpr (Div x y) = binOp (/) x y
Now we can use evalExpr to evaluate an expression in our mini-language with variable substitution. All the error handling if there's an undefined variable is done by the Maybe monad, and we were even able to cut down on repetition by making the environment argument implicit. This is all pretty standard for a simple expression DSL.
So for the fun bit, generating random expressions and (eventually) mutations. For this, we'll need System.Random. We want to be able to generate expressions of varying complexity, so we'll express it in rough depth. Since it's random, there is a chance that we'll get shorter or deeper trees than specified. This will probably be something that you'll want to tweak and tune to get your desired results. First, because I have the foresight of having written this code already, let's define two helpers for generating a random literal and a random variable:
randomLit, randomVar :: IO Expr
randomLit = Lit <$> randomRIO (-100, 100)
randomVar = Var <$> randomRIO ('x', 'z')
Nothing exciting here, we get doubles between -100 and 100, and up to 3 variables. Now we can generate large expression trees.
generateExpr :: Int -> IO Expr
-- When the depth is 1, return a literal or a variable randomly
generateExpr 1 = do
isLit <- randomIO
if isLit
then randomLit
else randomVar
-- Otherwise, generate a tree using helper
generateExpr n = randomRIO (0, 100) >>= helper
where
helper :: Int -> IO Expr
helper prob
-- 20% chance that it's a literal
| prob < 20 = randomLit
-- 10% chance that it's a variable
| prob < 30 = randomVar
-- 15% chance of Add
| prob < 45 = (+) <$> generateExpr (n - 1) <*> generateExpr (n - 1)
-- 15% chance of Sub
| prob < 60 = (-) <$> generateExpr (n - 1) <*> generateExpr (n - 1)
-- 15% chance of Mul
| prob < 75 = (*) <$> generateExpr (n - 1) <*> generateExpr (n - 1)
-- 15% chance of Div
| prob < 90 = (/) <$> generateExpr (n - 1) <*> generateExpr (n - 1)
-- 10% chance that we generate a possibly taller tree
| otherwise = generateExpr (n + 1)
The bulk of this function is just specifying the probabilities that a given node will be generated, and then generating the left and right nodes for each operator. We even got to use the normal arithmetic operators since we overloaded Num, how handy!
Now, remember that we can still pattern match on the constructors of this Expr type for other operations, such as replacing nodes, swapping them, or measuring the depth. For this, you just have to treat it as any other binary tree type in Haskell, except it has 2 leaf constructors and 4 node constructors. As for mutations, you'll have to write code that traverses this structure and chooses when to swap out nodes and what to swap them out with. It'll live within the IO monad since you'll be generating random values, but it shouldn't be too difficult. This structure should be pretty easy to extend as need be, such as if you wanted to add trig functions and exponentiation, you'd just need more constructors, more expressions in evalExpr, and the appropriate clauses in helper, along with some probability adjustments.
You can get the full code for this example here. Hope this helps you see how to formulate something like S-expressions in Haskell.
I use System.Random and System.Random.Shuffle to shuffle the order of characters in a string, I shuffle it using:
shuffle' string (length string) g
g being a getStdGen.
Now the problem is that the shuffle can result in an order that's identical to the original order, resulting in a string that isn't really shuffled, so when this happens I want to just shuffle it recursively until it hits a a shuffled string that's not the original string (which should usually happen on the first or second try), but this means I need to create a new random number generator on each recursion so it wont just shuffle it exactly the same way every time.
But how do I do that? Defining a
newg = newStdGen
in "where", and using it results in:
Jumble.hs:20:14:
Could not deduce (RandomGen (IO StdGen))
arising from a use of shuffle'
from the context (Eq a)
bound by the inferred type of
shuffleString :: Eq a => IO StdGen -> [a] -> [a]
at Jumble.hs:(15,1)-(22,18)
Possible fix:
add an instance declaration for (RandomGen (IO StdGen))
In the expression: shuffle' string (length string) g
In an equation for `shuffled':
shuffled = shuffle' string (length string) g
In an equation for `shuffleString':
shuffleString g string
= if shuffled == original then
shuffleString newg shuffled
else
shuffled
where
shuffled = shuffle' string (length string) g
original = string
newg = newStdGen
Jumble.hs:38:30:
Couldn't match expected type `IO StdGen' with actual type `StdGen'
In the first argument of `jumble', namely `g'
In the first argument of `map', namely `(jumble g)'
In the expression: (map (jumble g) word_list)
I'm very new to Haskell and functional programming in general and have only learned the basics, one thing that might be relevant which I don't know yet is the difference between "x = value", "x <- value", and "let x = value".
Complete code:
import System.Random
import System.Random.Shuffle
middle :: [Char] -> [Char]
middle word
| length word >= 4 = (init (tail word))
| otherwise = word
shuffleString g string =
if shuffled == original
then shuffleString g shuffled
else shuffled
where
shuffled = shuffle' string (length string) g
original = string
jumble g word
| length word >= 4 = h ++ m ++ l
| otherwise = word
where
h = [(head word)]
m = (shuffleString g (middle word))
l = [(last word)]
main = do
g <- getStdGen
putStrLn "Hello, what would you like to jumble?"
text <- getLine
-- let text = "Example text"
let word_list = words text
let jumbled = (map (jumble g) word_list)
let output = unwords jumbled
putStrLn output
This is pretty simple, you know that g has type StdGen, which is an instance of the RandomGen typeclass. The RandomGen typeclass has the functions next :: g -> (Int, g), genRange :: g -> (Int, Int), and split :: g -> (g, g). Two of these functions return a new random generator, namely next and split. For your purposes, you can use either quite easily to get a new generator, but I would just recommend using next for simplicity. You could rewrite your shuffleString function to something like
shuffleString :: RandomGen g => g -> String -> String
shuffleString g string =
if shuffled == original
then shuffleString (snd $ next g) shuffled
else shuffled
where
shuffled = shuffle' string (length string) g
original = string
End of answer to this question
One thing that might be relevant which I don't know yet is the difference between "x = value", "x <- value", and "let x = value".
These three different forms of assignment are used in different contexts. At the top level of your code, you can define functions and values using the simple x = value syntax. These statements are not being "executed" inside any context other than the current module, and most people would find it pedantic to have to write
module Main where
let main :: IO ()
main = do
putStrLn "Hello, World"
putStrLn "Exiting now"
since there isn't any ambiguity at this level. It also helps to delimit this context since it is only at the top level that you can declare data types, type aliases, and type classes, these can not be declared inside functions.
The second form, let x = value, actually comes in two variants, the let x = value in <expr> inside pure functions, and simply let x = value inside monadic functions (do notation). For example:
myFunc :: Int -> Int
myFunc x =
let y = x + 2
z = y * y
in z * z
Lets you store intermediate results, so you get a faster execution than
myFuncBad :: Int -> Int
myFuncBad x = (x + 2) * (x + 2) * (x + 2) * (x + 2)
But the former is also equivalent to
myFunc :: Int -> Int
myFunc x = z * z
where
y = x + 2
z = y * y
There are subtle difference between let ... in ... and where ..., but you don't need to worry about it at this point, other than the following is only possible using let ... in ..., not where ...:
myFunc x = (\y -> let z = y * y in z * z) (x + 2)
The let ... syntax (without the in ...) is used only in monadic do notation to perform much the same purpose, but usually using values bound inside it:
something :: IO Int
something = do
putStr "Enter an int: "
x <- getLine
let y = myFunc (read x)
return (y * y)
This simply allows y to be available to all proceeding statements in the function, and the in ... part is not needed because it's not ambiguous at this point.
The final form of x <- value is used especially in monadic do notation, and is specifically for extracting a value out of its monadic context. That may sound complicated, so here's a simple example. Take the function getLine. It has the type IO String, meaning it performs an IO action that returns a String. The types IO String and String are not the same, you can't call length getLine, because length doesn't work for IO String, but it does for String. However, we frequently want that String value inside the IO context, without having to worry about it being wrapped in the IO monad. This is what the <- is for. In this function
main = do
line <- getLine
print (length line)
getLine still has the type IO String, but line now has the type String, and can be fed into functions that expect a String. Whenever you see x <- something, the something is a monadic context, and x is the value being extracted from that context.
So why does Haskell have so many different ways of defining values? It all comes down to its type system, which tries really hard to ensure that you can't accidentally launch the missiles, or corrupt a file system, or do something you didn't really intend to do. It also helps to visually separate what is an action, and what is a computation in source code, so that at a glance you can tell if an action is being performed or not. It does take a while to get used to, and there are probably valid arguments that it could be simplified, but changing anything would also break backwards compatibility.
And that concludes today's episode of Way Too Much Information(tm)
(Note: To other readers, if I've said something incorrect or potentially misleading, please feel free to edit or leave a comment pointing out the mistake. I don't pretend to be perfect in my descriptions of Haskell syntax.)
I want to implement an imperative language interpreter in Haskell (for educational purposes). But it's difficult for me to create right architecture for my interpreter: How should I store variables? How can I implement nested function calls? How should I implement variable scoping? How can I add debugging possibilities in my language? Should I use monads/monad transformers/other techniques? etc.
Does anybody know good articles/papers/tutorials/sources on this subject?
If you are new to writing this kind of processors, I would recommend to put off using monads for a while and first focus on getting a barebones implementation without any bells or whistles.
The following may serve as a minitutorial.
I assume that you have already tackled the issue of parsing the source text of the programs you want to write an interpreter for and that you have some types for capturing the abstract syntax of your language. The language that I use here is very simple and only consists of integer expressions and some basic statements.
Preliminaries
Let us first import some modules that we will use in just a bit.
import Data.Function
import Data.List
The essence of an imperative language is that it has some form of mutable variables. Here, variables simply represented by strings:
type Var = String
Expressions
Next, we define expressions. Expressions are constructed from integer constants, variable references, and arithmetic operations.
infixl 6 :+:, :-:
infixl 7 :*:, :/:
data Exp
= C Int -- constant
| V Var -- variable
| Exp :+: Exp -- addition
| Exp :-: Exp -- subtraction
| Exp :*: Exp -- multiplication
| Exp :/: Exp -- division
For example, the expression that adds the constant 2 to the variable x is represented by V "x" :+: C 2.
Statements
The statement language is rather minimal. We have three forms of statements: variable assignments, while loops, and sequences.
infix 1 :=
data Stmt
= Var := Exp -- assignment
| While Exp Stmt -- loop
| Seq [Stmt] -- sequence
For example, a sequence of statements for "swapping" the values of the variables x and y can be represented by Seq ["tmp" := V "x", "x" := V "y", "y" := V "tmp"].
Programs
A program is just a statement.
type Prog = Stmt
Stores
Now, let us move to the actual interpreter. While running a program, we need to keep track of the values assigned to the different variables in the programs. These values are just integers and as a representation of our "memory" we just use lists of pairs consisting of a variable and a value.
type Val = Int
type Store = [(Var, Val)]
Evaluating expressions
Expressions are evaluated by mapping constants to their value, looking up the values of variables in the store, and mapping arithmetic operations to their Haskell counterparts.
eval :: Exp -> Store -> Val
eval (C n) r = n
eval (V x) r = case lookup x r of
Nothing -> error ("unbound variable `" ++ x ++ "'")
Just v -> v
eval (e1 :+: e2) r = eval e1 r + eval e2 r
eval (e1 :-: e2) r = eval e1 r - eval e2 r
eval (e1 :*: e2) r = eval e1 r * eval e2 r
eval (e1 :/: e2) r = eval e1 r `div` eval e2 r
Note that if the store contains multiple bindings for a variable, lookup selects the bindings that comes first in the store.
Executing statements
While the evaluation of an expression cannot alter the contents of the store, executing a statement may in fact result in an update of the store. Hence, the function for executing a statement takes a store as an argument and produces a possibly updated store.
exec :: Stmt -> Store -> Store
exec (x := e) r = (x, eval e r) : r
exec (While e s) r | eval e r /= 0 = exec (Seq [s, While e s]) r
| otherwise = r
exec (Seq []) r = r
exec (Seq (s : ss)) r = exec (Seq ss) (exec s r)
Note that, in the case of assignments, we simply push a new binding for the updated variable to the store, effectively shadowing any previous bindings for that variable.
Top-level Interpreter
Running a program reduces to executing its top-level statement in the context of an initial store.
run :: Prog -> Store -> Store
run p r = nubBy ((==) `on` fst) (exec p r)
After executing the statement we clean up any shadowed bindings, so that we can easily read off the contents of the final store.
Example
As an example, consider the following program that computes the Fibonacci number of the number stored in the variable n and stores its result in the variable x.
fib :: Prog
fib = Seq
[ "x" := C 0
, "y" := C 1
, While (V "n") $ Seq
[ "z" := V "x" :+: V "y"
, "x" := V "y"
, "y" := V "z"
, "n" := V "n" :-: C 1
]
]
For instance, in an interactive environment, we can now use our interpreter to compute the 25th Fibonacci number:
> lookup "x" $ run fib [("n", 25)]
Just 75025
Monadic Interpretation
Of course, here, we are dealing with a very simple and tiny imperative language. As your language gets more complex, so will the implementation of your interpreter. Think for example about what additions you need when you add procedures and need to distinguish between local (stack-based) storage and global (heap-based) storage. Returning to that part of your question, you may then indeed consider the introduction of monads to streamline the implementation of your interpreter a bit.
In the example interpreter above, there are two "effects" that are candidates for being captured by a monadic structure:
The passing around and updating of the store.
Aborting running the program when a run-time error is encountered. (In the implementation above, the interpreter simply crashes when such an error occurs.)
The first effect is typically captured by a state monad, the second by an error monad. Let us briefly investigate how to do this for our interpreter.
We prepare by importing just one more module from the standard libraries.
import Control.Monad
We can use monad transformers to construct a composite monad for our two effects by combining a basic state monad and a basic error monad. Here, however, we simply construct the composite monad in one go.
newtype Interp a = Interp { runInterp :: Store -> Either String (a, Store) }
instance Monad Interp where
return x = Interp $ \r -> Right (x, r)
i >>= k = Interp $ \r -> case runInterp i r of
Left msg -> Left msg
Right (x, r') -> runInterp (k x) r'
fail msg = Interp $ \_ -> Left msg
Edit 2018: The Applicative Monad Proposal
Since the Applicative Monad Proposal (AMP) every Monad must also be an instance of Functor and Applicative. To do this we can add
import Control.Applicative -- Otherwise you can't do the Applicative instance.
to the imports and make Interp an instance of Functor and Applicative like this
instance Functor Interp where
fmap = liftM -- imported from Control.Monad
instance Applicative Interp where
pure = return
(<*>) = ap -- imported from Control.Monad
Edit 2018 end
For reading from and writing to the store, we introduce effectful functions rd and wr:
rd :: Var -> Interp Val
rd x = Interp $ \r -> case lookup x r of
Nothing -> Left ("unbound variable `" ++ x ++ "'")
Just v -> Right (v, r)
wr :: Var -> Val -> Interp ()
wr x v = Interp $ \r -> Right ((), (x, v) : r)
Note that rd produces a Left-wrapped error message if a variable lookup fails.
The monadic version of the expression evaluator now reads
eval :: Exp -> Interp Val
eval (C n) = do return n
eval (V x) = do rd x
eval (e1 :+: e2) = do v1 <- eval e1
v2 <- eval e2
return (v1 + v2)
eval (e1 :-: e2) = do v1 <- eval e1
v2 <- eval e2
return (v1 - v2)
eval (e1 :*: e2) = do v1 <- eval e1
v2 <- eval e2
return (v1 * v2)
eval (e1 :/: e2) = do v1 <- eval e1
v2 <- eval e2
if v2 == 0
then fail "division by zero"
else return (v1 `div` v2)
In the case for :/:, division by zero results in an error message being produced through the Monad-method fail, which, for Interp, reduces to wrapping the message in a Left-value.
For the execution of statements we have
exec :: Stmt -> Interp ()
exec (x := e) = do v <- eval e
wr x v
exec (While e s) = do v <- eval e
when (v /= 0) (exec (Seq [s, While e s]))
exec (Seq []) = do return ()
exec (Seq (s : ss)) = do exec s
exec (Seq ss)
The type of exec conveys that statements do not result in values but are executed only for their effects on the store or the run-time errors they may trigger.
Finally, in the function run we perform a monadic computation and process its effects.
run :: Prog -> Store -> Either String Store
run p r = case runInterp (exec p) r of
Left msg -> Left msg
Right (_, r') -> Right (nubBy ((==) `on` fst) r')
In the interactive environment, we can now revisit the interpretation of our example program:
> lookup "x" `fmap` run fib [("n", 25)]
Right (Just 75025)
> lookup "x" `fmap` run fib []
Left "unbound variable `n'"
A couple of good papers I've finally found:
Building Interpreters by Composing Monads
Monad Transformers Step by Step - how incrementally build tiny interpreter using
monad transformers
How to build a monadic interpreter in one day
Monad Transformers and Modular Interpreters
So I'm writing a program which returns a procedure for some given arithmetic problem, so I wanted to instance a couple of functions to Show so that I can print the same expression I evaluate when I test. The trouble is that the given code matches (-) to the first line when it should fall to the second.
{-# OPTIONS_GHC -XFlexibleInstances #-}
instance Show (t -> t-> t) where
show (+) = "plus"
show (-) = "minus"
main = print [(+),(-)]
returns
[plus,plus]
Am I just committing a mortal sin printing functions in the first place or is there some way I can get it to match properly?
edit:I realise I am getting the following warning:
Warning: Pattern match(es) are overlapped
In the definition of `show': show - = ...
I still don't know why it overlaps, or how to stop it.
As sepp2k and MtnViewMark said, you can't pattern match on the value of identifiers, only on constructors and, in some cases, implicit equality checks. So, your instance is binding any argument to the identifier, in the process shadowing the external definition of (+). Unfortunately, this means that what you're trying to do won't and can't ever work.
A typical solution to what you want to accomplish is to define an "arithmetic expression" algebraic data type, with an appropriate show instance. Note that you can make your expression type itself an instance of Num, with numeric literals wrapped in a "Literal" constructor, and operations like (+) returning their arguments combined with a constructor for the operation. Here's a quick, incomplete example:
data Expression a = Literal a
| Sum (Expression a) (Expression a)
| Product (Expression a) (Expression a)
deriving (Eq, Ord, Show)
instance (Num a) => Num (Expression a) where
x + y = Sum x y
x * y = Product x y
fromInteger x = Literal (fromInteger x)
evaluate (Literal x) = x
evaluate (Sum x y) = evaluate x + evaluate y
evaluate (Product x y) = evaluate x * evaluate y
integer :: Integer
integer = (1 + 2) * 3 + 4
expr :: Expression Integer
expr = (1 + 2) * 3 + 4
Trying it out in GHCi:
> integer
13
> evaluate expr
13
> expr
Sum (Product (Sum (Literal 1) (Literal 2)) (Literal 3)) (Literal 4)
Here's a way to think about this. Consider:
answer = 42
magic = 3
specialName :: Int -> String
specialName answer = "the answer to the ultimate question"
specialName magic = "the magic number"
specialName x = "just plain ol' " ++ show x
Can you see why this won't work? answer in the pattern match is a variable, distinct from answer at the outer scope. So instead, you'd have to write this like:
answer = 42
magic = 3
specialName :: Int -> String
specialName x | x == answer = "the answer to the ultimate question"
specialName x | x == magic = "the magic number"
specialName x = "just plain ol' " ++ show x
In fact, this is just what is going on when you write constants in a pattern. That is:
digitName :: Bool -> String
digitName 0 = "zero"
digitName 1 = "one"
digitName _ = "math is hard"
gets converted by the compiler to something equivalent to:
digitName :: Bool -> String
digitName x | x == 0 = "zero"
digitName x | x == 1 = "one"
digitName _ = "math is hard"
Since you want to match against the function bound to (+) rather than just bind anything to the symbol (+), you'd need to write your code as:
instance Show (t -> t-> t) where
show f | f == (+) = "plus"
show f | f == (-) = "minus"
But, this would require that functions were comparable for equality. And that is an undecidable problem in general.
You might counter that you are just asking the run-time system to compare function pointers, but at the language level, the Haskell programmer doesn't have access to pointers. In other words, you can't manipulate references to values in Haskell(*), only values themselves. This is the purity of Haskell, and gains referential transparency.
(*) MVars and other such objects in the IO monad are another matter, but their existence doesn't invalidate the point.
It overlaps because it treats (+) simply as a variable, meaning on the RHS the identifier + will be bound to the function you called show on.
There is no way to pattern match on functions the way you want.
Solved it myself with a mega hack.
instance (Num t) => Show (t -> t-> t) where
show op =
case (op 6 2) of
8 -> "plus"
4 -> "minus"
12 -> "times"
3 -> "divided"