Re: [Haskell-beginners] Picking apart getLine

OK, so if the user enters "ABC\n" then the following happens: x <- getChar --A x <- getChar --B x <- getChar --B x <- getChar --\n levels are different contexts? levels of recursion. then when recursion ends get 'A' : 'B' : 'C' : [] So does the (x:xs) syntax mean this? I thought of (x:xs) as 'A' : ['B','C'] But can the xs mean simply the rest? ie (x:xs) in this case x = 'A' and xs represents the rest 'B' : 'C' : [] Is that maybe the way to think of it? I suppose it does. The confusing bit is how all the x's after the first one (ie after 'A') are represented in returm (x:xs). But I am now thinking that ['B','C'] is actually the same as 'B' : 'C' : [] and all that the 'B' and 'C' and also the last [] are the xs part of (x:xs) On 7 January 2014 17:15, David McBride wrote:

You have the idea. The x is fetched with getChar, then it sits in that context until the return is executed.

So the x is sitting there and getLine' is called. It makes its own x via getChar, then maybe getLine' is called again. Each getLine' sits there with its own version of x until finally the last getLine' get's a \n, and then returns a []. Then the whole thing unwinds by prepending x to [], then x to [x], then another x to [x,x], until there are no more x's to return, and you have the whole string.

Hopefully that paragraph makes sense.

On Tue, Jan 7, 2014 at 11:54 AM, Angus Comber wrote:

...

Before looking at getLine, I can understand this:

getnumber :: IO Int getnumber = do x <- getChar if isDigit x then return (ord x - ord '0') else return 0

OK, it is not a very useful function but at least I understand it. return is required so that the function returns an IO Int.

But I don't understand this:

getLine' :: IO String getLine' = do x <- getChar if x == '\n' then return [] else do xs <- getLine' return (x:xs)

I can understand what will happen if a user enters a newline (only). return [] brings the empty list into the monadic world.

But what is happening if x is not a newline?

xs <- getLine' will recursively call getChar and retrieve another character from the input stream. But it will do this BEFORE the return (x:xs) - so what is happening to all the head elements in the list - the x element?

It is difficult to picture in my mind how this is working. I understand recursion but this looks tricky.

Can someone help me work this out?

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners