[Haskell-beginners] merge error

[Hemanth Gunda]

Hi Trent,

*why first == first@(x:xs), especially weather the variable declarations
are considered names for the same thing. *

first@(x:xs) is meant to be read as below:

"first as (x:xs)".

This is syntactic sugar that gives you the flexibility to address the
entire list as "first", the first element (head) of the list as "x" and the
rest of the list (tail) as xs.

If you don't want to use @, you can write the last case in your program as:

merge first second | (head first) <= (head second)     = (head first) :
merge (tail first) second

                   | otherwise  = (head second): merge first (tail second)

instead of

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

                                 | otherwise  = y : merge first ys

Relevant SO question :
https://stackoverflow.com/questions/1153465/what-does-the-symbol-mean-in-reference-to-lists-in-haskell

*why [x] /= xs *

when you call merge as:

> merge first second

from merge type signature,

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

 we know it takes  two lists of type a, belonging to typeclass Ord (where
<, >  has meaning)

first & second checked for being a list & whatever is present inside the
list is expected to be of typeclass Ord.

Case A: If you wrote merge [x] [] = [x]

[x] is expected to be a list of Ord
which means x is expected to be Ord.

Case B: if you called merge x [] = x

x is expected to be a list of Ord.

When you call merge [4,5] []

In Case A, it implies "4,5" is a member of Ord, as this is x. Which it
isn't (as 3 <"4,5" has no meaning). Therefore merge [x] [] = [x] isn't
executed here, the program goes looking for other cases & doesn't find any
that satisfies these input types & leads to the error you encountered.

In Case B, it imples [4,5] is a list of Ords, which corresponds to the
correct type. Therefore this statement is executed.

Regards, Hemanth


On Fri, May 18, 2018 at 11:34 AM trent shipley <trent.shipley at gmail.com>
wrote:

> 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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