Do-notation and the list monad-CodePudding

I am learning Haskell.

I am trying to find elements of a list as which sum to elements of a list bs, returning the elements as a tuple:

findSum2 :: [Int] -> [Int] -> [(Int,Int,Int)]
findSum2 as bs = [(a, a', b) | a <- as, a' <- as, b <- bs, a   a' == b]

The code works. But in order to learn Haskell, I'm trying to rewrite it as do-notation:

findSum2 :: [Int] -> [Int] -> [(Int,Int,Int)]
findSum2 as bs = do
  a  <- as
  a' <- as 
  b  <- bs
  if a   a' == b then return (a, a', b) else return ()

The typechecker then complains at me:

   • Couldn't match type ‘()’ with ‘(Int, Int, Int)’
      Expected type: [(Int, Int, Int)]
        Actual type: [()]

In all fairness, I knew it would. But since I can't skip the else clause in Haskell, what should I put in the return statement in the else clause?

Thanks.

CodePudding user response：

You can not return the unit (()), since that means that the return (a, a', b) and the return () have different types: the first one is [(Int, Int, Int)], whereas the latter is [()].

What you can do is use an empty list in case the guard fails, so:

findSum2 :: [Int] -> [Int] -> [(Int,Int,Int)]
findSum2 as bs = do
  a  <- as
  a' <- as 
  b  <- bs
  if a   a' == b then return (a, a', b) else []

CodePudding user response：

You must return something of the correct type in the else clause. It could be the empty list [], or one of the abstracted values like mzero or empty.

Or you could remove the if expression and use the guard function.

import Control.Monad (guard)

findSum2 :: [Int] -> [Int] -> [(Int,Int,Int)]
findSum2 as bs = do
  a  <- as
  a' <- as 
  b  <- bs
  guard (a   a' == b)
  return (a, a', b)

With this implementation you could now also generalize your function signature to:

findSum2 :: MonadPlus m => m Int -> m Int -> m (Int, Int, Int)