Meaning of a semicolon in lambda expression

Question

Type:

data Command a = Command String (a -> IO a) 

Function:

iofunc_ :: String -> (a -> IO ()) -> Command a
iofunc_ s f = Command s (\x -> do f x ; return x)

What does the semicolon do in the lambda expression (\x -> do f x ; return x)?

Answer 1

They just separate the two expression f x and return x in do notation. In fact these all are equivalent in your case:

iofunc_ s f = Command s (\x -> do f x ; return x)

iofunc_ s f = Command s (\x -> do {f x ; return x})

iofunc_ s f = Command s (\x -> do f x
                                  return x)

iofunc_ s f = Command s (\x -> f x >> return x)