Re: [Haskell-beginners] merge error

Thanks to all. I used Mukesh's suggestion. I am still not clear on: why [x] /= xs why first == first@(x:xs), especially weather the variable declarations are considered names for the same thing. On Thu, May 17, 2018 at 10:39 PM Hemanth Gunda wrote:

Hi Trent,

This works:

merge:: Ord a => [a] -> [a] -> [a] merge [] [] = [] merge x [] = x merge [] y = y merge first@(x:xs) second@(y:ys) | x <= y = x : merge xs second | otherwise = y : merge first ys

Difference in the lines

merge x [] = x merge [] y = y

As the input is of type [a] where a belongs to typeclass Ord, you must pass x instead of [x].

[x] would work if you tried merge [4] []. but will fail if you tried merge [4,5] []. because "4,5" isn't of type a.

Regards, Hemanth

On Fri, May 18, 2018 at 10:51 AM trent shipley wrote:

...

The below produces an error. And I am very proud that I could use the GHCi debugging tools to get this far.

merge [] [] works.

merge [1] [] works.

I don't know why the failing example fails. It should return:

[4,5]

Help to unstuck is appreciated.

:step merge [4,5] []

*** Exception: ex6_8.hs:(12,1)-(16,66): Non-exhaustive patterns in function merge

Given:

merge :: Ord a => [a] -> [a] -> [a]

merge [] [] = []

merge [x] [] = [x]

merge [] [y] = [y]

merge first@(x:xs) second@(y:ys) | x <= y = x : merge xs second

| otherwise = y : merge first ys

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