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@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@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@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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