On Wed, Apr 3, 2013 at 5:01 AM, Angus Comber <anguscom...@gmail.com> wrote:
> In the (x:xs) : just delimits each element. so x is the first element. Why > can I not print by using x? > > Also xs is of what type? list of values? So does this mean x is an element > and xs must be of type list? Confused... > Actually, you just answered yourself. x is an element, xs is a list. (++) combines lists, so to insert your element using (++) you need to make it a list. [x] is a list containing your element x and nothing else. Another way to do it is to do the same as the pattern match, but in this case that's kinda ugly: firstLetter all@(x:xs) = "The first letter of " ++ all ++ " is " ++ (x : " otherbit ") ++ xs or firstLetter all@(x:xs) = "The first letter of " ++ all ++ " is " ++ (x : []) ++ " otherbit " ++ xs (note that "is of type list" is incomplete; list of *what*? In this case, list of Char. Haskell String is just [Char] (list of Char), which is highly convenient but a bit slow in real programs that manipulate a lot of String-s.) -- brandon s allbery kf8nh sine nomine associates allber...@gmail.com ballb...@sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net
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