I am very new to Haskell and I need to make a working calculator what will give answers to expressions like: 2+3*(5+12)
I have something that manages to calculate more or less but I have a problem with order of operations. I have no idea how to do it. Here is my code:
import Text.Regex.Posix
import Data.Maybe
oblicz :: String -> Double
oblicz str = eval (Nothing, None) $ map convertToExpression $ ( tokenize str )
eval :: (Maybe Double,Expression)->[Expression]->Double
eval (Nothing, _) ((Variable v):reszta) = eval (Just v, None) reszta
eval (Just aktualnyWynik, None) ((Operator o):reszta) = eval ((Just aktualnyWynik), (Operator o)) reszta
eval (Just aktualnyWynik, (Operator o)) ((Variable v):reszta) = eval (Just $ o aktualnyWynik v , None) reszta
eval (aktualnyWynik, operator) (LeftParenthesis:reszta)
= eval (aktualnyWynik, operator) ((Variable (eval (Nothing, None) reszta)):(getPartAfterParentheses reszta))
eval (Just aktualnyWynik, _) [] = aktualnyWynik
eval (Just aktualnyWynik, _) (RightParenthesis:_) = aktualnyWynik
data Expression = Operator (Double->Double->Double)
| Variable Double
| LeftParenthesis
| RightParenthesis
| None
tokenize :: String -> [String]
tokenize expression = getAllTextMatches(expression =~ "([0-9]+|\\(|\\)|\\+|-|%|/|\\*)" :: AllTextMatches [] String)
convertToExpression :: String -> Expression
convertToExpression "-" = Operator (-)
convertToExpression "+" = Operator (+)
convertToExpression "*" = Operator (*)
convertToExpression "/" = Operator (/)
convertToExpression "(" = LeftParenthesis
convertToExpression ")" = RightParenthesis
convertToExpression variable = Variable (read variable)
getPartAfterParentheses :: [Expression] -> [Expression]
getPartAfterParentheses [] = []
getPartAfterParentheses (RightParenthesis:expressionsList) = expressionsList
getPartAfterParentheses (LeftParenthesis:expressionsList) = getPartAfterParentheses (getPartAfterParentheses expressionsList)
getPartAfterParentheses (expression:expressionsList) = getPartAfterParentheses expressionsList
I thought maybe I could create two stacks - one with numbers and one with operators. While reading the expression, I could push numbers on one stack and operators on another. When it is an operator I would check if there is something already on the stack and if there is check if I should pop it from the stack and do the math or not - just like in onp notation.
Unfortunately, as I said, I am VERY new to haskell and have no clue how to go about writing this.
Any hints or help would be nice :)
Pushing things on different stacks sure feels very much a prcedural thing to do, and that's generally not nice in Haskell. (Stacks can be realised as lists, which works quite nice in a purely functional fashion. Even real mutable state can be fine if only as an optimisation, but if more than one object needs to be modified at a time then this isn't exactly enjoyable.)
The preferrable way would be to build up a tree representing the expression.
type DInfix = Double -> Double -> Double -- for readability's sake
data ExprTree = Op DInfix ExprTree ExprTree
| Value Double
Evaluating this tree is basically evalTree (Op c t1 t2) = c (evalTree t1) (evalTree t2), i.e. ExprTree->Double right away.
To build the tree up, the crucial point: get the operator fixities right. Different operators have different fixity. I'd put that information in the Operator field:
type Fixity = Int
data Expression = Operator (Double->Double->Double) Fixity
| ...
which then requires e.g.
...
convertToExpression "+" = Operator (+) 6
convertToExpression "*" = Operator (*) 7
...
(Those are the fixities that Haskell itself has for the operators. You can :i + in GHCi to see them.)
Then you'd build the tree.
toExprTree :: [Expression] -> ExprTree
Obvious base case:
toExprTree [Variable v] = Value v
You might continue with
toExprTree (Variable v : Operator c _ : exprs) = Op c (Value v) (toExprTree exprs)
But that's actually not right: for e.g. 4 * 3 + 2 it would give 4 * (3 + 2). We actually need to bring the 4 * down the remaining expressions tree, as deep as the fixities are lower. So the tree needs to know about that as well
data ExprTree = Op DInfix Fixity ExprTree ExprTree
| Value Double
mergeOpL :: Double -> DInfix -> Fixity -> ExprTree -> ExprTree
mergeOpL v c f t#(Op c' f' t' t'')
| c > c' = Op c' f' (mergeOpL v c f t') t''
mergeOpL v c f t = Op c f (Value v) t
What remains to be done is handling parentheses. You'd need to take a whole matching-brackets expression and assign it a tree-fixity of, say tight = 100 :: Fixity.
As a note: such a tokenisation - manual parsing workflow is pretty cumbersome, regardless how nicely functional you do it. Haskell has powerful parser-combinator libraries like parsec, which take most of the work and bookkeeping off you.
What you need to solve this problem is the Shunting-yard Algorithm of Edsger Dijstra as described at http://www.wcipeg.com/wiki/Shunting_yard_algorithm. You can see my implementation at the bottom of this file.
If you are limiting your self to just +,-,*,/ you can also solve the problem using the usual trick in most intro to compiler examples simply parsing into two different non-terminals, ofter called term and product to build the correct tree. This get unwieldy if you have to deal with a lot of operators or they are user defined.
Related
I have 2 lists Expressions and bindings (id = Expr), and trying to replace each expression with its equivalent from the bindings list in a new list called newE, where Expression = id ..
Initially, I have only one expression:
eq (div (add 2 7) (sub 5 2)) 3
I want to replace each identifier in this expression with its equivalent from the bindings list, so I tried to split this expression into a list of strings and removed brackets, to separate each identifier ..
This is how I tried to implement it:
newE = [\x -> getExp(b) | x <- eStr, b <- bs, x == getId(b)]
where es = getExpressions (prog)
bs = getBindings (prog)
-- Extracting expression into a list of strings
-- and removing brackets
eStr = map (delete ')')(map (delete ')')
(map (delete '(') (split " " (show es))))
newE' = join " " newE
Is this correct?
Now I'm getting errors that newE returns [t0 -> Expr] while it's supposed to return Expr only, why is that?
and the join function is expecting a [Char] .. while its type in the Data.List.Utils documentation is:
join :: [a] -> [[a]] -> [a]
so, isn't it supposed to accept any type not just list of characters? or did it get confused with a 'join' from another library?
I searched the libraries I've imported, but they don't have a join.
Any help to resolve these errors and modify the code to do what it's supposed to do?
Thank you
Since you asked, here is a sketch of the conventional approach:
Convert the string expression into a Expr value, e.g.
add 2 7 -> App (App (Var "add") (Var "2")) (Var "7")
Write a function lookupBinding to lookup a binding for a Symbol:
lookupBinding :: [Binding] -> Symbol -> Maybe Expr
Write a substitute function to substitute binding definitions into an expression:
substitute :: [Binding] -> Expr -> Expr
It will go something like this:
substitute bindings (App e1 e2) = App e1' e2'
where e1' = substitute bindings e1
e2' = substitute bindings e2
substitute bindings (Var sym) = ... lookupBinding sym ...
substitute bindings (Lam sym exp) = ... substitute bindings' exp ...
Your list comprehension body returns the function \x -> getExp b. That's why the compiler says it's returning a function type.
This is the file I am trying to load:
import Data.List (foldl')
import Text.Printf (printf)
import Data.Char (ord)
--data IntParsedStr = Int | ParsingError
--data ParsingError = ParsingError String
asInt :: String -> Either String Integer
asInt "" = Left "Empty string"
asInt xs#(x:xt) | x == '-' = either Left (Right . ((-1) *)) (asInt' xt)
| otherwise = either Left Right (asInt' xs)
where asInt' :: String -> Either String Integer
asInt' "" = Left "No digits"
asInt' xs = foldl' step (Right 0) xs
where step :: Either String Integer -> Char -> Either String Integer
step (Left e) _ = Left e
step (Right ac) c = either Left (Right . (10 * ac + ) . fromIntegral) (charAsInt c)
charAsInt :: Char -> Either String Int
charAsInt c | between (ord '0') (ord c) (ord '9') = Right $ ord c - ord '0'
| otherwise = Left $ printf "'%c' is not a digit" c
checkPrecision str = either error ((str == ). show) (asInt str)
between :: Ord t => t -> t -> t -> Bool
between a b c = a <= b && b <= c
It loads without any problem in ghci but in hugs I get this error:
ERROR "folds.hs":17 - Syntax error in expression (unexpected `)')
Line 17 is the last in the definition of asInt function
Edit:
Hi!
I recently founded that this is in fact a known hugs issue as said in this question
where there is a link to the Hugs 98 Users Guide where says that
Legal expressions like (a+b+) and (a*b+) are rejected.
I believe this is a bug in Hugs, not a liberality of GHC. The Haskell 98 report (appropriate in the context of Hugs usage) says
Syntactic precedence rules apply to sections as follows. (op e) is legal if and only if (x op e) parses in the same way as (x op (e)); and similarly for (e op). For example, (*a+b) is syntactically invalid, but (+a*b) and (*(a+b)) are valid. Because (+) is left associative, (a+b+) is syntactically correct, but (+a+b) is not; the latter may legally be written as (+(a+b)).
I interpret that as allowing (10 * ac + ) since both (*) and (+) are left associative, and indeed (*) has higher precedence.
As pointed out in the comments, ((10 * ac) + ) is accepted by both, so is a workaround.
Interestingly, this isn't listed in the Hugs vs Haskell 98 page, so maybe Mark P. Jones reads this section of the report differently to me. Certainly I can forgive him for this; Gofer implemented constructor classes long before Haskell allowed them, and Hugs is still faster than GHCi at compilation, and still gives better error messages.
for a homework assignment, a subtask is to make the arithmetic functions (+), (-), (*) and div showable.
We're solved the rest of the assignment, but we're stuck here. Right now we're using the solution to this question here to distinguish between the operations:
showOp op = case op 3 3 of
6 -> "plus"
0 -> "minus"
9 -> "times"
1 -> "divide"
_ -> "undefined"
However, this strikes me as kind of ugly as things like showOp (\a b -> a * 3 - y) yield "plus".
Is there any way to better distinguish between the operators?
We are using winhugs atm with the appropriate switches -98 +o in order to be able to use the needed extensions.
Edit:
As requested, the actual assignment has to do with Arrays (specifically Array Int (Int -> Int -> Int)). It has to do with generating arrays of operators that fulfill certain conditions.
The assignment states:
Make the data type Array Int (Int->Int-Int) an Instance of Show. The arithmetic operations from the previous exercises should be represented as "plus", "minus", "times" and "div".
thx for any help in advance
Use induction :)
{-# LANGUAGE FlexibleInstances #-}
instance Eq (Int-> Int -> Int) where
f == g = induce f g where
base = 1
n = 2
induce f g = and [f 1 n' == g 1 n' | n' <- [base, n, n+1]]
instance Show (Int-> Int -> Int) where
show a = showOp a where
showOp op = case lookup op ops of
Just a -> a
otherwise -> "undefined"
ops = [((+),"plus")
,((-),"minus")
,((*),"times")
,(div,"divide")]
Output:
*Main> (\a b -> a * 3 - b) :: (Int->Int->Int)
undefined
Reading through Why It’s Nice to be Quoted, in section 3 there's an example of splicing a variable identifier in a quasiquote.
subst [:lam | $exp:e1 $exp:e2 |] x y =
let e1' = subst e1 x y
e2' = subst e2 x y
in
[:lam | $exp:e1' $exp:e2' |]
I see why the recursive calls to subst are done outside the [:lam| ... |], it's because the function antiVarE in section 3.2 builds a TH.varE out of the variable name.
My question is how much work would be required to support arbitrary expression splices beyond just a variable name?
For example:
subst [:lam | $exp:e1 $exp:e2 |] x y =
[:lam | $exp:(subst e1 x y) $exp:(subst e2 x y) |]
Answering my own question for posterity.
Turns out it was quite simple. Using the parseExp function in haskell-src-meta package I was able to easily convert a string to AST fragment.
In the original paper, aside from the parser changes required to capture an expression string between parentheses, the antiExpE function could be rewritten as such.
antiExpE :: Exp -> Maybe TH.ExpQ
antiExpE (AE v) =
case parseExp v of
Right exp -> Just . return $ exp
Left _ -> Nothing
antiExpE = Nothing
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"