Philip Cote
So a week into Haskell, I still seem to be not “getting” it which is kind of weird in my case. I came in with knowledge of a lot of functional ideas from using them in Javascript and Python. Or at least I thought I knew them.
5 chapters into "Learn You a Haskell", I admit it's not really sinking in for me even after typing in and running all the examples. I acknowledge that I don't know jack.
This is going to be a somewhat long mail, in which I'm attempting to give you an introduction of my own. Be brave and read it. =) Well, despite popular belief JavaScript and Python are not anywhere near functional. JS just uses callbacks and passes them around through higher order functions, but that's just the regular event-handling imperative style. There is really nothing functional about that. Functional style in JS starts when you start using things like 'each' from jQuery, i.e. you're not just defining event handlers. You are actually passing /behavior/ to a function. This is basically what a functional language takes to its conclusion. You have an action (like putStrLn "Hello world"), and you want to perform it n times. Let me give this example in both JavaScript and Haskell: function timesHello(n) { while (n > 0) { print("Hello world"); n--; } } timesHello 0 = return () timesHello n = putStrLn "Hello world" >> nTimes (n - 1) But why should we be constrained to printing hello world? On the other hand, how could we abstract it? The complicated answer is to use a complicated OO solution, but functional programmers can do a lot better: function times(n, action) { while (n > 0) { action(); n--; } } times 0 action = return () times n action = action >> times (n - 1) action Unfortunately JavaScript has quite a heavy syntax for anonymous functions, which makes this style a lot less useful: times(3, function() { print("Hello world"); }); But in Haskell we can write this elegantly: 3 `times` putStrLn "Hello world" This is the very fundamental idea. Since Haskell programs are lazily evaluated you can stretch this idea to entirely new dimensions, where the line between data and control structures disappears. A common statement is that lists in Haskell correspond to 'for' loops in other languages, which is quite true, but can only be seen when you ask yourself, "how would I solve this?". Let me give you an example: You are searching for the smallest square natural number that has a digit sum greater than 20. First we need a function to calculate digit sums. This would be a 'for' loop in an imperative language, wouldn't it? In Haskell we do something different: Our function starts iterating division by 10: digitSum = iterate (`div` 10) digitSum 357 = [357, 35, 3, 0, 0, 0, ... Then we want to get rid of the zeroes. Starting with the first zero all following numbers will be zeroes, so we stop before the first zero: digitSum = takeWhile (> 0) . iterate (`div` 10) digitSum 357 = [357, 35, 3] As you can see we have already used two higher order functions that take behavior as arguments. Next we need to take everything modulo 10: digitSum = map (`mod` 10) . takeWhile (> 0) . iterate (`div` 10) digitSum 357 = [7, 5, 3] Finally we take the sum: digitSum = sum . map (`mod` 10) . takeWhile (> 0) . iterate (`div` 10) digitSum 357 = 15 Ok, let's solve the original problem. Again this would be another 'for' loop in other languages, but there is a Haskell solution as well: The smallest square natural number... "smallest natural" sounds a lot like we want to search from 0 upwards, so let's just capture our search space as a list: [0..] = [0, 1, 2, 3, 4, ... But we want squares: map (^2) [0..] = [0, 1, 4, 9, 16, ... And we want only the numbers with a digit sum greater than 20: filter (\x -> digitSum x > 20) . map (^2) $ [0..] The natural number 44944 is a square, and its digit sum is greater than 20, so this list is certainly nonempty. So there must be a smallest number in that list, which will be just the first entry, because the list is sorted by construction. In other words, it's safe to use 'head': head . filter (\x -> digitSum x > 20) . map (^2) $ [0..] And that gives us the solution 1849, which has a digit sum of 22 and meets our specification.
Any ideas or exercises that might help me along?
Yes. Pick a small task and solve it. Write code. If you get stuck, just visit us at #haskell on Freenode or ask here. I hope this helped. Greets, Ertugrul -- Not to be or to be and (not to be or to be and (not to be or to be and (not to be or to be and ... that is the list monad.