On Mon, Oct 17, 2011 at 04:18:05PM +0100, Hugo Ferreira wrote:
Hello,
I came across the following code:
ngrams'' :: Int -> [a] -> [[a]] ngrams'' n l = do t <- Data.List.tails l l <- [take n t] Control.Monad.guard (length l == n) return l
and tried to use the ">>=" operator in order to figure out how Monads work. I came up with:
test l = (Data.List.tails l)
= (\t -> [take 2 t]) = (\l -> if (length l == 2) then [l] else [])
Questions: 1. How can I use Control.Monad.guard directly in "test l"
test l = (Data.List.tails l) >>= \t -> [take 2 t] >>= \l -> Control.Monad.guard (length l == 2) >> return l The rule is that x <- foo desugars to foo >>= \x -> ... and blah desugars to blah >> ... One thing that might have been tripping you up is your extra parentheses around the lambda expressions. If you have
= (\l -> ...) foo...
the l does not scope over foo... so you cannot mention it. Instead what you want is
= \l -> ... foo...
so the lambda expression is actually \l -> ... >> foo..., that is, it includes *everything* after the \l -> ... and not just the stuff on that line.
2. Related issue: how can create a List monad so that I can input "l" argument of "test l" directly?
I don't understand this question. Can you give an example of what you are trying to do? -Brent