[Haskell-beginners] merge error

[Ut Primum]

[x] means a list containing the element x

xs in general is only a variable named xs, and in particular could be a
list; in this case, if you write
first@(x:xs)
you mean that first is a list, its "head" is x and xs is its "tail", i.e.
all the elements of the list first following x.

For example if the arguments of merge are [1,2,4] and [3,6], then
first = [1,2,4]
x=1
xs = [2,4]
second=[3,6]
y=3
ys=[6]

Note that [1,2,4] could not be pattern-matchd with [x]. [6] could (because
contains just one element)

Il ven 18 mag 2018, 08:04 trent shipley <trent.shipley at gmail.com> ha
scritto:

> 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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