[Haskell-beginners] merge error

[trent shipley]

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 <hemanth.420 at gmail.com>
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 <trent.shipley at gmail.com>
> 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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