Let’s say that you have a list of values and you needed to check if any of those values satisfied some condition.
This can be solved easily with Haskell’s any function, but let’s say you didn’t have that.
Here’s an alternative method using foldr.
any' :: (a -> Bool) -> [a] -> Bool
any' p list = foldr ((||) . p) False list
any' (== 'x') ['x','y','z']
-- True
any' even [1,3,5,7]
-- FalseTo understand how this works, you’ve got to wrap your head around two non-trivial functions for haskell beginners: foldr and ..
Dot
. takes two functions and composes them to create a new function. This is best seen by way of example.
If you know that length counts items in a list and even tests if a number is even, then when someone asks you to write a function that determines if the number of characters in a string is even, it should be little more than these two words put together some way…
-- types refresher:
--
-- String = [Char]
-- length :: [a] -> Int
-- even :: Int -> Bool
--
-- correct, but ugly
stringIsEven :: String -> Bool
stringIsEven s = even (length s)
-- better, but too long
stringIsEven s = even $ length s
-- perfect
stringIsEven = even . lengthRemember, $ is function application while . is function composition.
I think that gives you a general idea for how it works. Now let’s translate that to the specific example at hand:
(||) is just another function. Think about that for a minute. Welcome to haskell.
(||) takes two Bools and returns a Bool; True if either one of its arguments is True.
Curious how (||) is defined in haskell’s Prelude?
Wow.
Haskell’s laziness means there’s no special tricks needed to make if statements “short circuit”. Haskell won’t evaluate the second expression if the first is True because it’s simply never needed.
OK, back to our function.
p is provided as the first argument to our any' function and we know that it’s type is (a -> Bool). This means it has to be a test that will check a value and return True or False.
So, what might the type of ((||) . p) be?
This composed function (and you’ve really got to think of it as one function) will take some value, a as its first argument. It will apply p to it which gives an intermediate Bool. That Bool is then passed through as the first argument to (||).
(||), having gotten its first argument already, only needs one more argument. Since it’s not supplied by anything else, it’s now required as an argument to the composed function.
-- suppose p is already defined like so:
p :: Char -> Bool
p = (== 'z')
-- types refresher:
--
-- p :: Char -> Bool
-- (||) :: Bool -> Bool -> Bool
--
((||) . p) 'z' False
-- True
((||) . p) 'x' True
-- True
-- ((||) . p) :: Char -> Bool -> BoolEasy, right?
Folds
The next crazy function is foldr. A fold in the general sense is a way to reduce a list.
If you’ve got a list of items, a reducing function, and some initial value, then a fold is the process of using these three things to reduce the list to a single value.
This can be seen in the type of foldr. A great deal of information can be learned in haskell by simply taking a look at types; that’s why haddocks are so invaluable.
foldr :: (a -> b -> b) -- ^ a reducing function
-> b -- ^ some initial value
-> [a] -- ^ a list of items
-> b -- ^ the resultant single valueTake care to note the type of the reducing function.
It must accept as its first argument the same type as your list of items contains and as its second argument the same type as your initial value.
Its result is also the same type as your initial value. This is important because the initial value and the result of the previous application of foldr must be the same type if we want the required recursion to be type safe.
Often, the types a and b are the same (as in sum' explained below), but this is not required.
foldr and foldl are different in the direction of the fold: folding to the right or folding to the left. In some cases this doesn’t matter, in others it does.
Let’s look at a folding sum as a concrete example:
sum' :: [Int] -> Int
sum' xs = foldr (+) 0 xs
sum' [1,2,3,4,5]
-- 15
-- how's it work?
foldr (+) 0 [1,2,3,4,5]
-- 15
-- the reducing function is applied with its second argument as the
-- initial value
result = (+) 1 0 -- 1 + 0 = 1
-- now we apply the same function but use the result of the previous
-- application as the new initial value and act on the next element
result' = (+) 2 1 -- 2 + 1 = 3
-- rinse and repeat until all elements are used up
result'' = (+) 3 3 -- 3 + 3 = 6
result''' = (+) 4 6 -- 4 + 6 = 10
result'''' = (+) 5 10 -- 5 + 10 = 15
((((0 + 1) + 2) + 3) + 4) + 5
-- 15Here’s another breakdown with the recursion explicitly shown rather than the values it represents:
foldr (+) 0 [1,2,3,4,5]
result = (+) 1 0
result' = (+) 2 $ (+) 1 0
result'' = (+) 3 $ (+) 2 $ (+) 1 0
result''' = (+) 4 $ (+) 3 $ (+) 2 $ (+) 1 0
result'''' = (+) 5 $ (+) 4 $ (+) 3 $ (+) 2 $ (+) 1 0
-- 15This is an easy example where you can see clearly how things work out. In our case it’s a bit more complex, but the principle is the same:
-- assume p is defined like so
p :: Char -> Bool
p = (== 'b')
foldr ((||) . p) False ['a','b','c']
-- True
-- value breakdown:
result = ((||) . p) 'a' False -- (== 'b') 'a' || False = False
result' = ((||) . p) 'b' False -- (== 'b') 'b' || False = True DING!
result'' = ((||) . p) 'c' True -- (== 'b') 'c' || True = True
-- recursion breakdown:
result = ((||) . p) 'a' False
result'' = ((||) . p) 'b' $ ((||) . p) 'a' False -- <- this is the only
result''' = ((||) . p) 'c' $ ((||) . p) 'b' $ ((||) . p) 'a' False -- line ever evaulated
-- TrueSo the whole thing reduces to True, just as we’d expect.
Why?
This was a really slow and deliberate explanation. I did it this way because I had a real-world situation where I had to come to this exact understanding to solve a problem. OK, not really to solve some dire problem per say, but to do something I wanted to do in an elegant way…
I wanted to walk you all through it from the start only so someone not-so-familiar with haskell might a) see its beauty and b) actually understand the single line of code I’m going to show you in a few more paragraphs.
Sorry.
In my window manager of choice, XMonad, there is a means to test a window’s property and take some action depending on the result.
The simplest example is move windows with class “firefox” to the “web” workspace.
Easy.
There’s also a means to OR rules like these together.
With this, I can say move windows with class “firefox” OR title “chrome” to the “web” workspace.
The two functions (=?) and (<||>) behave exactly like their normal (==) and (||) counterparts. They’re just lifted into a Query Monad. This is a concept you don’t need to comprehend right now, just know that there’s no elegant way to apply any in this context.
That made it difficult to write a simple function: matchAny that could be a test if any of a window’s properties (class, title, name, or role) matched the test string.
Now the any' exercise isn’t looking so unrealistic, is it?
-- types refresher:
--
-- any :: (a -> Bool) -> [a] -> Bool
--
-- we need the same thing, just lifted to the "Query" context:
-- liftAny :: (a -> Query Bool) -> [a] -> Query Bool
--
-- our any reimplimentation from ealier:
any' p list = foldr ((||) . p) False list
-- almost identical:
liftAny p list = foldr ((<||>) . p) (return False) listreturn False is simply the lifted version of False just like (=?) is the lifted version of (==)…
Now my manage hooks can leverage a list comprehension for a much more concise and readable rule.
matchAny :: String -> Query Bool
matchAny s = liftAny (=? s) [className, title, name, role]
myManageHook = composeAll [ matchAny s --> action | (s, action) <- myActions ]
where
myActions = [ ("rdesktop" , doFloat )
, ("Xmessage" , doCenterFloat )
, ("Gmrun" , doCenterFloat )
, ("Uzbl" , doShift "2-web" )
, ("Uzbl-core" , doShift "2-web" )
, ("Chromium" , doShift "2-web" )
, ("irssi" , doShift "3-chat")
]Finally
So why is this post about laziness?
That statement “short circuits”. That’s only possible because of lazy evaluation.