haskell happy - "Couldn't match expected type"

Question

I have some problems to understand error messages of the parser generator system happy for haskell.

For instance this code

{
module Test_parser where

import System.IO
import Test_lexer

}

%name parse

%tokentype { Token }
%error { parseError }

%token

    true                { TokenTrue posn }
    false               { TokenFalse posn }
    number              { TokenInteger posn $$ }

%%

LiteralExpr :
        true            {LiteralTrue ()}
    | false         {LiteralFalse ()}
    | number        {LiteralInteger () $1}

{

data Expression n =
    LiteralTrue n
    | LiteralFalse n
    | LiteralInteger n Integer
        deriving (Eq, Show)

parseError xs = error "Parse error"

main = do 
  inStr <- getContents
  let parseTree = parse (alexScanTokens inStr)  
  putStrLn $ show(parseTree)
}

works fine.

But if I do something like that

....

LiteralExpr :
    true            {LiteralTrue ()}
    | false         {LiteralFalse ()}
    | Inte          {$1}

Inte :
    number          {LiteralInteger () $1}


{

data Expression n =
    LiteralTrue n
    | LiteralFalse n
    | Inte n
        deriving (Eq, Show)

data Inte n =
    LiteralInteger n Integer
        deriving (Eq, Show)

....

I will get a huge error message. So it would be very nice if you could help me to understand it.

thanks in advance

score 2 · Answer 1 · answered Oct 07 '12 at 20:36

2

Seems like you're confusing data constructors and type constructors. Inte in Expression is a data constructor that happened to have the same name as (and isn't otherwise related to) the Inte datatype that you define below.

You should have something like

data Expression n =
    LiteralTrue n
    | LiteralFalse n
    | LiteralInt (Inte n)

and modify your code appropriately (add LiteralInt where needed).

answered Oct 07 '12 at 20:36

Roman Cheplyaka

37,738
7
72
121

Sorry, I don't get it. Maybe you are right but I still get error messages. This does not work :".. LiteralExpr : true {LiteralTrue ()} | false {LiteralFalse ()} | Inte {LiteralInt (Inte () $1)} Inte : number {LiteralInteger () $1} { data Expression n = LiteralTrue n | LiteralFalse n | LiteralInt (Inte n) deriving (Eq, Show) data Inte n = LiteralInteger n Integer deriving (Eq, Show) .." Could you please modify the given code example to run it. Maybe I get it after seeing the correct version. ;) – jimmyt Oct 07 '12 at 23:16

score 0 · Answer 2 · answered Oct 08 '12 at 08:15

ok, this is the way it goes:

...
LiteralExpr :
    true            {LiteralTrue ()}
    | false         {LiteralFalse ()}
    | Inte          {LiteralInt $1}

Inte :
    number          {LiteralInteger () $1}
...

and

...
data Expression n =
    LiteralTrue n
    | LiteralFalse n
    | LiteralInt (Inte n)
        deriving (Eq, Show)

data Inte n =
    LiteralInteger n Integer
        deriving (Eq, Show)
...

haskell happy - "Couldn't match expected type"

2 Answers2