How do i write the following function with the >>= operator

Question

How do I write this function using the >>= operator?

parseNumber2 :: Parser LispVal
parseNumber2 = do x <- many1 digit
                  return $ (Number . read) x

Answer 1

A straightforward desugaring of the do-notation gives

parseNumber2 :: Parser LispVal
parseNumber2 = many1 digit >>= (return . Number . read)

but the more idiomatic way is to use fmap or the equivalent <$> operator from Control.Applicative

parseNumber2 = Number . read <$> many1 digit

To desugar do-notation:

1. Flip any <- bindings over to the right side and add >>= and a lambda abstraction

   do x <- a
      y <- b
      ...
   

   becomes

   a >>= \x ->
   b >>= \y ->
   ...
   

2. For any non-binding forms, add a >> on the right:

   do a
      b
      ...
   

   becomes

   a >>
   b >>
   ...
   

3. Leave the last expression alone.

   do a
   

   becomes

   a
   

Applying these rules to your code, we get

parseNumber2 =
    many1 digit >>= \x -> 
    return $ (Number . read) x

Do some simplifications

parseNumber2 = many1 digit >>= \x -> (return . Number . read) x
parsenumber2 = many1 digit >>= (return . Number . read)

Now, for any monad, fmap or <$> can be defined as

f <$> x = x >>= (return . f)

Use this to get the idiomatic form

parseNumber2 = Number . read <$> many1 digit