On Tuesday 07 June 2011, 14:30:29, Christopher Howard wrote:
In a reply to an earlier question, someone told me that "do" expressions are simply syntactic sugar that expand to expressions with the >> and
= operators. I was trying to understand this (and the Monad class) better.
I see in my own experiments that this...
main = do ln1 <- getLine putStrLn ln1
translates to this:
main = getLine >>= \ln1 -> putStrLn ln1
Shorter: getLine >>= putStrLn An expression `\x -> foo x' is equivalent to `foo', using a lambda doesn't make things clearer in such cases, only if something more complicated is done with the argument(s).
However, what does this translate to?:
main = do ln1 <- getLine ln2 <- getLine putStrLn (ln1 ++ ln2)
That desugars via getLine >>= \ln1 -> (do { ln2 <- getLine; putStrLn (ln1 ++ ln2); }) to getLine >>= \ln1 -> (getLine >>= \ln2 -> putStrLn (ln1 ++ ln2)) Note that the outer parentheses aren't necessary. do val <- foo bar baz val desugars to foo >>= \val -> do { bar; baz val; } and the you can desugar the right hand side recursively.