31 Aug
2012
31 Aug
'12
8:16 a.m.
my mistake, that's not right. But you should try to use use something like this: intercalate :: (Eq t) => [t] -> [t] -> [t] intercalate _ [] = [] intercalate [] _ = [] intercalate (x:xs) (y:ys) = x : y : intercalate xs ys with the last line, you imply that your lists have at least one element (through pattern matching). By the recursive call, you will enter this 'non-exhaustive' case. Also, instead of writing *intercalate (x:xs) (y:ys)* >> * | xt == [] = []* >> * | yt == [] = []* >> * | otherwise = x : y : intercalate xs ys* * where xt=(x:xs)* >> * yt=(y:ys)* for defining xt and yt, you could write: intercalate xt@(x:xs) yt@(y:ys) greets On 08/31/2012 02:06 PM, Gary Klindt wrote: > Hi, > > you don't proof for empty lists, only for empty tails: (x:[]) > > > On 08/31/2012 02:01 PM, Ezequiel Hernan Di Giorgi wrote: >> First i want to thank all the persons who responded me yesterday to help >> me. Thanks! I am so happy with with your friendliness. >> So i have other beginners question: >> >> Now i want a improved version of my* intercalate*. Now i want to call a >> function with two [t][t] and obtain another one which have only >> even elements, even because: >> >> - [1,2,3,3,4][6] and the ouput [1,6] >> - [1,2,3,4][5,6,7] output [1,5,2,6,3,7] >> >> I tried it: >> >> *intercalate :: (Eq t) => [t] -> [t] -> [t]* >> *intercalate (x:xs) (y:ys)* >> * | xt == [] = []* >> * | yt == [] = []* >> * | otherwise = x : y : intercalate xs ys* >> * where xt=(x:xs)* >> * yt=(y:ys)* >> >> but i get nice error >> >> **Main> intercalate [1][6]* >> *[1,6*** Exception: baby.hs:(2,1)-(5,51): Non-exhaustive patterns in >> function intercalate* >> * >> * >> **Main> * >> >> (yes...the file's name is baby.hs) >> >> Thanks in advance! (: (: (: >> (: (: >> >> >> >> _______________________________________________ >> Beginners mailing list >> Beginners@haskell.org >> http://www.haskell.org/mailman/listinfo/beginners >>