I'm very new to Elm (a couple of weeks) and am working through exercises on exercism.io. I see a lot of potential in this language and am enjoying the learning process. Writing about what I learn usually helps me understand it better. Thus, this article was born.
The exercise is simple: identify leap years with Elm. So, given a year, determine whether that year is a leap year or not. The rules are as follows:
leap years are on: - every year that is evenly divisible by 4 - except every year that is evenly divisible by 100 - unless the year is also evenly divisible by 400
Let's start by writing some functions for checking these divisibility rules.
divisibleBy4 year = year % 4 == 0 divisibleBy100 year = year % 100 == 0 divisibleBy400 year = year % 400 == 0
Now for the boolean expression...
(divisibleBy4 year) && (not (divisibleBy100 year) || (divisibleBy400 year))
Cool. So now we can put the pieces together and build our function.
But first, a function signature!
We want a function that takes an integer (
Int) and returns a boolean (
isLeapYear : Int -> Bool
Now, we can add a
let-in expression to create a local function scope for our
isLeapYear year = let divisibleBy4 y = y % 4 == 0 divisibleBy100 y = y % 100 == 0 divisibleBy400 y = y % 400 == 0 in (divisibleBy4 year) && (not (divisibleBy100 year) || (divisibleBy400 year))
This could actually be much better.
First off, we can make
divisibleByX a lot more useful by having it take two arguments rather than one: the number to divide and the number to divide by.
divisibleBy n y = y % n == 0
Nice, but there's more. One of Elm's core libraries, "Basics" has a function called
rem with the following type signature:
rem : Int -> Int -> Int
According to the docs,
it does this:
Find the remainder after dividing one number by another. e.g. rem 11 4 == 3
Is this exactly what we need? Yes, this is exactly what we need.
divisibleBy n y = rem y n == 0
Noice. Time to refactor.
isLeapYear : Int -> Bool isLeapYear y = let divisibleBy n y = rem y n == 0 in (divisibleBy 4 y) && (not (divisibleBy 100 y) || (divisibleBy 400 y))
That boolean expression looks like it could be simplified. There's another interesting boolean operator in the Basics package...
xor : Bool -> Bool -> Bool The exclusive-or operator. True if exactly one input is True.
Cool beans. Let's refactor our solution.
If we read our requirements again, this problem actually lends itself quite well to
We want all years divisible by 4 EXCEPT the ones divisible by 100.
xor (divisibleBy 4 year) (divisibleBy 100 year) || (divisibleBy 400 year)
This leaves us with a third iteration of:
isLeapYear : Int -> Bool isLeapYear year = let divisibleBy n y = rem y n == 0 in xor (divisibleBy 4 year) (divisibleBy 100 year) || (divisibleBy 400 year)
I've not even scratched the surface of this powerful language. Getting into the FRP mindset has been a challenge but I can already see the benefits of thinking in this paradigm. While the solution I've put together over the last few paragraphs gets the job done, there is surely a better way...