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 ha scritto:

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

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