Using ParSec integer function to give a Int?

Question

I need to get an Int type from the integer function with Parsec. My code at the moment is

aTerm =  parens aExpression
     <|> liftM GetV identifier
     <|> liftM N integer

Where the type of N is

N    :: Num a => a -> Expr a

The error I am getting is

Shane.hs:227:18:
    Couldn't match expected type `Int' with actual type `Integer'
    Expected type: Text.Parsec.Prim.ParsecT String u0 Identity Int
      Actual type: Text.Parsec.Prim.ParsecT String u0 Identity Integer
    In the second argument of `liftM', namely `integer'
    In the second argument of `(<|>)', namely `liftM N integer'
Failed, modules loaded: none.

Is it possible to extract an Int type here? Using fromInteger some way? Changing from Int to Integer isn't an option.

Depending on your use case, it might be necessary to check whether the parsed `Integer` is in the `Int` range, and if not let the parse fail. — Daniel Fischer, Nov 25 '12 at 19:03

score 3 · Accepted Answer · answered Nov 25 '12 at 17:23

3

This should work.

aTerm =  parens aExpression
     <|> liftM GetV identifier
     <|> liftM (N . fromInteger) integer

answered Nov 25 '12 at 17:23

dave4420

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Using ParSec integer function to give a Int?

1 Answers1