In what does this simple do notation desugar to?

Question

I'm using Aeson and Network.HTTP. I was able to encode a json and print it on the screen doing the following:

getCode :: String -> IO ResponseCode
getCode url = simpleHTTP req >>= getResponseCode
    where req = getRequest url

main :: IO ()
main = do 
    x <- get "http://jsonplaceholder.typicode.com/todos/1"
    let y = encode x
    B.putStrLn y

However I do not understand what this do expression desugars to. Something like this:

get "http://jsonplaceholder.typicode.com/todos/1" >>= (?)

what should be in the ??

I only know how to desugar this:

do { x1 <- action1
   ; x2 <- action2
   ; mk_action3 x1 x2 }

to this

action1 >>= (\ x1 -> action2 >>= (\ x2 -> mk_action3 x1 x2 ))

bu the way, what is an action? https://en.m.wikibooks.org/wiki/Haskell/do_notation does not explain very precisely.

Answer 1

get "http://jsonplaceholder.typicode.com/todos/1" >>= (\x -> let y = encode x in B.putStrLn y)

Answer 2

put (x :: s) is a State s "action".

put :: s -> State s () is a function from s type values to State s "actions", an "action" constructor.

putStrLn "Hi" is an IO action. putStrLn :: String -> IO () is a function from Strings to IO actions, an "action" constructor.

IO is a Monad. State s is a Monad.

an "action" is any value of type M a where M is a Monad.