Why does type ParsecT have an argument "u"?

Question

The documentation for the parsec package states that the argument u used to transfer some user state through monadic computation. But the same functionality can be achieved by basing the ParsecT monad transformer on State monad. Therefore, if my parser is not workable, I do not need u in general, but I need to set it to () using the parse. What is the rationale for adding optional state support to ParsecT ?

Answer 1

Because the ParsecT s () (State st) a parser behaves differently than the Parsec s st Identity a parser when it comes to backtracking:

• The user state is reset when parsec tries to use the alternative after a failure that does not use input.
• But the base Monad m does not back down; all effects that occurred on the way to the final analysis result are saved.

Consider the following example:

 {-# LANGUAGE FlexibleContexts #-} module Foo where import Control.Applicative import Control.Monad.State import Text.Parsec.Prim hiding ((<|>), State(..)) import Text.Parsec.Error (ParseError) tick :: MonadState Int m => ParsecT s Int m () tick = do lift $ modify (+1) modifyState (+1) tickTock :: MonadState Int m => ParsecT s Int m () tickTock = (tick >> empty) <|> tick -- | run a parser that has both user state and an underlying state monad. -- -- Example: -- >>> run tickTock -- (Right 1,2) run :: ParsecT String Int (State Int) () -> (Either ParseError Int, Int) run m = runState (runParserT (m >> getState) initUserState "-" "") initStateState where initUserState = 0 initStateState = 0 

As you can see, the main state monad registered two ticks (from both alternatives that were tested) while the user state of the Parsec monastic transformer only kept successful.

Answer 2

ParsecT already has its own state: position and input parsing: http://haddocks.fpcomplete.com/fp/7.8/20140916-162/parsec/Text-Parsec-Prim.html#t:State

As leftaroundabout noted, this is probably due to optimization.