Re: [Haskell-beginners] can I have a case statement with a pattern matching

I'm nitpicking here, but it can be improved a bit doubleEveryOther :: [Integer] -> [Integer] doubleEveryOther [] = [] doubleEveryOther [x] = [x] doubleEveryOther (x:y:xs) = x : 2 * y : doubleEveryOther (xs) 2015-05-14 8:48 GMT+02:00 Nishant :

This will double very second digit in list. To do from right, just reverse the list first and then apply doubleEverySecond and then reverse back.

doubleEveryOther :: [Integer] -> [Integer]

doubleEveryOther [] = [] doubleEveryOther (x : []) = [x] doubleEveryOther (x : y : []) = x : (2 * y) : [] doubleEveryOther (x:y:xs) = x : (2 * y) : doubleEveryOther (xs)

On Wed, May 13, 2015 at 10:22 PM, Roelof Wobben wrote:

...

Hello,

I try to make make the assignment of CIS194 where I must multiply every second item from the right by 2

So I thought this could be working :

-- | Convert a digit to a reversed list for example 123 becomes [3,2,1] revtoDigits :: Integer -> [Integer] revtoDigits number | number <= 0 = [] | number > 0 = number`mod` 10 : revtoDigits (number `div` 10)

-- | Convert a digit to a list for example 123 becomes [1,2,3] toDigits :: Integer -> [Integer] toDigits number | number <= 0 = [] | number > 0 = toDigitsAcc [] number

toDigitsAcc:: [Integer] -> Integer -> [Integer] toDigitsAcc acc number | number <= 0 = acc | number > 0 = toDigitsAcc ((number `mod` 10) : acc) (number `div` 10)

doubleEveryOther :: [Integer] -> [Integer] doubleEveryOther list = case (length list) `mod` 2 of 0 -> doubleEveryOther [] = [] doubleEveryOther (x:y:xs) = x*2 : y : xs (length list) `mod` 2 of 0 -> doubleEveryOther [] = [] doubleEveryOther (x:y:xs) = x : y*2 : xs

-- | The main entry point. main = print $ doubleEveryOther [1,2,3]

but apperantly you cannot have pattern matching after a case statement because I get a error on the = of a empty list

Roelof

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