Understanding filterM

Question

Consider

filterM (\\x -> [True, False]) [1, 2, 3]

I just cannot understand the magic that Haskell does with this filterM use ca

Answer 1

This is pretty straightforward, after we've put it all down on paper (someone smart once gave this advice: don't try doing it all in your head, put it all on paper!):

filterM          :: (Monad m) => (a -> m Bool) -> [a] -> m [a]
filterM _ []     =  return []
filterM p (x:xs) =  do { flg <- p x
                       ; ys  <- filterM p xs
                       ; return (if flg then x:ys else ys) }

-- filterM (\x -> [True, False]) [1, 2, 3]
g [x,y,z] = filterM (\x -> [True, False]) (x:[y,z])
          = do {
                 flg <- (\x -> [True, False]) x
               ; ys  <- g [y,z]
               ; return ([x | flg] ++ ys) }
         = do {
                flg <- [True, False]               -- no `x` here!
              ; ys  <- do { flg2 <- (\x -> [True, False]) y
                          ; zs  <- g [z]
                          ; return ([y | flg2] ++ zs) }
              ; return ([x | flg] ++ ys) }
         = do {
                flg  <- [True, False]
              ; flg2 <- [True, False]
              ; zs   <- do { flg3 <- (\x -> [True, False]) z
                           ; s  <- g []
                           ; return ([z | flg3] ++ s) }
              ; return ([x | flg] ++ [y | flg] ++ zs) }
         = do {
                flg  <- [True, False]
              ; flg2 <- [True, False]
              ; flg3 <- [True, False]
              ; s    <- return []
              ; return ([x | flg] ++ [y | flg2] ++ [z | flg3] ++ s) }

The unnesting of the nested do blocks follows from the Monad laws.

Thus, in pseudocode:

    filterM (\x -> [True, False]) [1, 2, 3]
    =
      for flg in [True, False]:    -- x=1
          for flg2 in [True, False]:     -- y=2
              for flg3 in [True, False]:     -- z=3
                  yield ([1 | flg] ++ [2 | flg2] ++ [3 | flg3])