Hi,
between the arguments of a function,
ys@(y:yss)
would mean that we call that argument ys, we call its head y and its tail
yss. In your case the tail is replaced by _ , because is never used inside
the function.
Il mer 27 gen 2021, 07:02 Lawrence Bottorff
I'm looking at this https://wiki.haskell.org/99_questions/Solutions/8 and wondering how the "as pattern" actually is working
compress (x:ys@(y:_)) | x == y = compress ys | otherwise = x : compress ys compress ys = ys
I'm sure it's just some version of my stab at eliminating consecutive duplicates in a list
compress :: Eq a => [a] -> [a] compress [] = [] compress [x] = [x] compress (x:y:xs) = if x == y then compress (y:xs) else x : compress (y:xs)
only smarter. Could someone walk me through the (x:ys@(y:_)) part?
LB
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners