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Semicolon value in lambda expression

A 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 a semicolon do in a lambda expression (\x -> do f x ; return x)

?

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They just separate the two expressions f x

and return x

in notation. In fact, they are all 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)
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A semicolon anywhere is equivalent to changing the indented line at the same level as the previous valid expression.



I saw this while looking at how indentation works ( https://en.wikibooks.org/wiki/Haskell/Indentation ).

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