[Haskell-cafe] Parsec question

[Levi Stephen]

Dave Tapley wrote:
> I find it's good for the soul to remember what the do notation is 
> doing for us.
>
> Also I'm with Einstein on "You do not really understand something
> unless you can explain it to your grandmother"  :)
>
> Personally I think (in this instance) your three 'Parser a' functions
> read nicer as:
>
>> primary = (identifier >>= (return . PrimaryIdentifier)) <|> 
>> (stringLiteral >>= (return . PrimaryLiteral))
>> identifier = (many1 letter) >>= (return . Identifier)
>> stringLiteral = (char '\'') >> (manyTill anyChar (char '\'')) >>= 
>> (return . StringLiteral)
Thanks, I tried to do this, but got stuck. I was missing the 'return .'
>
> Looking at them in this form Tomek's point should seem clear now,
> especially when we look at the type signature for liftM:
>
> liftM :: Monad m => (a1 -> r) -> m a1 -> m r
>
> So we can (marginally) shorten down to:
>
>> primary = (liftM PrimaryIdentifier identifier) <|> (liftM 
>> PrimaryLiteral stringLiteral)
>> identifier = liftM Identifier (many1 letter)
>> stringLiteral = liftM StringLiteral ((char '\'') >> (manyTill anyChar 
>> (char '\'')))
I had initially tried following the types through of liftM as I didn't 
get Tomek's approach straight away. It did make a bit of sense, but with 
the extra step of getting rid of the do notation made it clearer.

>
>
> You might like:
> http://syntaxfree.wordpress.com/2006/12/12/do-notation-considered-harmful/ 
>
Thanks for the link. I have been trying to avoid do notation, as I think 
it helps my understanding of what's really going on with monads. In this 
case though, I got stuck when I left out the 'return .' and resorted to 
do notation because it worked :(

Levi
lstephen.wordpress.com
>