
[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
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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