Go line by line, except the last one:
a) When you see the pattern "a <- b", just swap it with "b >>= \s ->";
b) When you see the pattern "b", put a ">>" at the end of the line.
In your case:
| h = (return 1 :: IO Integer) >>= \a ->
| return 2 >>= \b ->
| return (a + b + 1)
An example when you have a line without an assignment ("<-"), this:
| h = do a <- (return 1 :: IO Integer)
| b <- (return 2)
| putStrLn "Hello, world!"
| return (a + b + 1)
Turns into:
| h = (return 1 :: IO Integer) >>= \a ->
| return 2 >>= \b ->
| putStrLn "Hello, world!" >>
| return (a + b + 1)
2012/9/4 Christopher Howard
What does the following do expression translate into? (I.e., using >>= operator and lambda functions.)
code: -------- h = do a <- (return 1 :: IO Integer) b <- (return 2) return (a + b + 1) --------