Re: [Haskell-beginners] Monad operators: multiple parameters

Also I don't really know how "let"s inside a do block desugars but I'm guessing it's simple substitution so for example the "main" function below probably translates to: main3 = let suffix = "suffix" in getLine >>= \x -> return (map toUpper x ++ suffix) -deech On Tue, Jun 7, 2011 at 11:58 AM, aditya siram wrote:

In your example, yes the two would produce the same result, but the difference with using "let" inside of a do-block is that you have access to variables captured inside the block. So for example if you wanted to get input from the user, capitalize it and append a suffix , the following functions are equivalent :    import Data.Char(toUpper)    main = let suffix = "suffix" in               do                   x <- getLine                   let upperCase = map toUpper x                   return (upperCase ++ suffix)    main2 = let suffix = "suffix" in                do                    x <- getLine                    return (map toUpper x ++ suffix)

Notice that in the "let upperCase ..." line in "main" uses "x" which was defined within the do-block, you would not have access to "x" outside the do-block. For reading simplicity when the whole function is one do-block I usually stick the "let"s inside the block.

-deech

On Tue, Jun 7, 2011 at 11:42 AM, Christopher Howard wrote:

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On 06/07/2011 04:54 AM, Mats Rauhala wrote:

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On 04:30 Tue 07 Jun     , Christopher Howard wrote:

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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

However, what does this translate to?:

main = do ln1 <- getLine           ln2 <- getLine           putStrLn (ln1 ++ ln2)

It translates to:

main = getLine >>= \ln1 -> getLine >>= \ln2 -> putStrLn (ln1 ++ ln2)

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Thank you Fischer and Rauhala for your prompt and helpful responses.

One final point of clarification, on a related note: am I correct in reasoning that...

main = do let x = expression          /remainder/

is the equivalent of...

main = let x = expression in /expanded remainder/

so that, for example...

main = ln1 <- getline       let ln2 = "suffix"       putStrLn (ln1 ++ ln2)

would translate to...

main = let ln2 = "suffix" in        getLine >>= \ln1 -> putStrLn (ln1 ++ ln2)

This above example works, but I want to be sure I haven't misunderstood any important technical points.

-- frigidcode.com theologia.indicium.us

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