Haskell - Happy - "No instance ..." error - parsing

Question

I'm trying to get familiar with Happy parser generator for Haskell. Currently, I have an example from the documentation but when I compile the program, I get an error.
This is the code:
{
module Main where
import Data.Char
}
%name calc
%tokentype { Token }
%error { parseError }
%token
let { TokenLet }
in { TokenIn }
int { TokenInt $$ }
var { TokenVar $$ }
'=' { TokenEq }
'+' { TokenPlus }
'-' { TokenMinus }
'*' { TokenTimes }
'/' { TokenDiv }
'(' { TokenOB }
')' { TokenCB }
%%
Exp : let var '=' Exp in Exp { \p -> $6 (($2,$4 p):p) }
| Exp1 { $1 }
Exp1 : Exp1 '+' Term { \p -> $1 p + $3 p }
| Exp1 '-' Term { \p -> $1 p - $3 p }
| Term { $1 }
Term : Term '*' Factor { \p -> $1 p * $3 p }
| Term '/' Factor { \p -> $1 p `div` $3 p }
| Factor { $1 }
Factor
: int { \p -> $1 }
| var { \p -> case lookup $1 p of
Nothing -> error "no var"
Just i -> i }
| '(' Exp ')' { $2 }
{
parseError :: [Token] -> a
parseError _ = error "Parse error"
data Token
= TokenLet
| TokenIn
| TokenInt Int
| TokenVar String
| TokenEq
| TokenPlus
| TokenMinus
| TokenTimes
| TokenDiv
| TokenOB
| TokenCB
deriving Show
lexer :: String -> [Token]
lexer [] = []
lexer (c:cs)
| isSpace c = lexer cs
| isAlpha c = lexVar (c:cs)
| isDigit c = lexNum (c:cs)
lexer ('=':cs) = TokenEq : lexer cs
lexer ('+':cs) = TokenPlus : lexer cs
lexer ('-':cs) = TokenMinus : lexer cs
lexer ('*':cs) = TokenTimes : lexer cs
lexer ('/':cs) = TokenDiv : lexer cs
lexer ('(':cs) = TokenOB : lexer cs
lexer (')':cs) = TokenCB : lexer cs
lexNum cs = TokenInt (read num) : lexer rest
where (num,rest) = span isDigit cs
lexVar cs =
case span isAlpha cs of
("let",rest) -> TokenLet : lexer rest
("in",rest) -> TokenIn : lexer rest
(var,rest) -> TokenVar var : lexer rest
main = getContents >>= print . calc . lexer
}
I'm getting this error:
[1 of 1] Compiling Main ( gr.hs, gr.o )
gr.hs:310:24:
No instance for (Show ([(String, Int)] -> Int))
arising from a use of `print'
Possible fix:
add an instance declaration for (Show ([(String, Int)] -> Int))
In the first argument of `(.)', namely `print'
In the second argument of `(>>=)', namely `print . calc . lexer'
In the expression: getContents >>= print . calc . lexer
Do you know why and how can I solve it?

Answer 1

If you examine the error message
No instance for (Show ([(String, Int)] -> Int))
arising from a use of `print'
it's clear that the problem is that you are trying to print a function. And indeed, the value produced by the parser function calc is supposed to be a function which takes a lookup table of variable bindings and gives back a result. See for example the rule for variables:
{ \p -> case lookup $1 p of
Nothing -> error "no var"
Just i -> i }
So in main, we need to pass in a list for the p argument, for example an empty list. (Or you could add some pre-defined global variables if you wanted). I've expanded the point-free code to a do block so it's easier to see what's going on:
main = do
input <- getContents
let fn = calc $ lexer input
print $ fn [] -- or e.g. [("foo", 42)] if you wanted it pre-defined
Now it works:
$ happy Calc.y
$ runghc Calc.hs <<< "let x = 1337 in x * 2"
2674