Trivial parsec example produces a type error

Question

I'm trying to get this trivial parsec code to compile

import Text.Parsec
simple = letter

but I keep getting this error

No instance for (Stream s0 m0 Char)
  arising from a use of `letter'
Possible fix: add an instance declaration for (Stream s0 m0 Char)
In the expression: letter
In an equation for `simple': simple = letter

I disagree that this is a duplicate. Yes, they concern the same basic issue, but that question is general, where this one is specific. Somebody searching for this specific instance of the problem doesn't need to know the full details of what the restriction is, they just need specific recommendations of how to resolve the problem - particularly, they need to know what the correct type signature for such a function would be. — Jules

fuz fuz · Accepted Answer · 2011-07-17T10:34:54

I think you have ran against the monomorphism restriction. This restriction means: If a variable is declared with no explicit arguments, its type has to be monomorphic. This forces the typechecker to pick a particular instance of Stream, but it can't decide.

There are two ways to fight it:

Give simple an explicit signature:

simple :: Stream s m Char => ParsecT s u m Char
simple = letter

Disable the monorphism restriction:

{-# LANGUAGE NoMonomorphismRestriction #-}
import Text.Parsec
simple = letter

See What is the monomorphism restriction? for more information on the monomorphism restriction.

Trivial parsec example produces a type error

1 Answers