Yet Another Haskell Tutorial/Language basics/Solutions
It binds more tightly; actually, function application binds more tightly than anything else. To see this, we can do something like:
Prelude> sqrt 3 * 3 5.19615
If multiplication bound more tightly, the result would have been 3.
Pairs, Triples and More
snd (fst ((1,'a'),"foo")). This is because first we want to take the first half the tuple:
(1,'a') and then out of this we want to take the second half, yielding just
If you tried
fst (snd ((1,'a'),"foo")) you will have gotten a type error. This is because the application of
snd will leave you with
fst "foo". However, the string "foo" isn't a tuple, so you cannot apply
fst to it.
Simple List Functions
map Char.isLower "aBCde"
length (filter Char.isLower "aBCde")
foldr max 0 [5,10,2,8,1].
You could also use
foldl. The foldr case is easier to explain: we replace each cons with an application of
max and the empty list with 0. Thus, the inner-most application will take the maximum of 0 and the last element of the list (if it exists). Then, the next-most inner application will return the maximum of whatever was the maximum before and the second-to-last element. This will continue on, carrying to current maximum all the way back to the beginning of the list.
In the foldl case, we can think of this as looking at each element in the list in order. We start off our "state" with 0. We pull off the first element and check to see if it's bigger than our current state. If it is, we replace our current state with that number and the continue. This happens for each element and thus eventually returns the maximal element.
fst (head (tail [(5,'b'),(1,'c'),(6,'a')]))
Source Code Files
We can define a fibonacci function as:
fib 1 = 1 fib 2 = 1 fib n = fib (n-1) + fib (n-2)
We could also write it using explicit if statements, like:
fib n = if n == 1 || n == 2 then 1 else fib (n-1) + fib (n-2)
Either is acceptable, but the first is perhaps more natural in Haskell.
We can define:
And then type out code:
mult a 0 = 0 mult a 1 = a mult a b = if b < 0 then 0 - mult a (-b) else a + mult a (b-1)
Note that it doesn't matter that of and we do the recursion on. We could just as well have defined it as:
mult 0 b = 0 mult 1 b = b mult a b = if a < 0 then 0 - mult (-a) b else b + mult (a-1) b
We can define
my_map f  =  my_map f (x:xs) = f x : my_map f xs
Recall that the
my_map function is supposed to apply a function
f to every element in the list. In the case that the list is empty, there are no elements to apply the function to, so we just return the empty list.
In the case that the list is non-empty, it is an element
x followed by a list
xs. Assuming we've already properly applied
xs, then all we're left to do is apply
x and then stick the results together. This is exactly what the second line does.
The code below appears in
Numbers.hs. The only tricky parts are the recursive calls in
module Main where import System.IO main = do nums <- getNums putStrLn ("The sum is " ++ show (sum nums)) putStrLn ("The product is " ++ show (product nums)) showFactorials nums getNums = do putStrLn "Give me a number (or 0 to stop):" num <- getLine if read num == 0 then return  else do rest <- getNums return ((read num :: Int):rest) showFactorials  = return () showFactorials (x:xs) = do putStrLn (show x ++ " factorial is " ++ show (factorial x)) showFactorials xs factorial 1 = 1 factorial n = n * factorial (n-1)
The idea for
getNums is just as spelled out in the hint. For
showFactorials, we consider first the recursive call. Suppose we have a list of numbers, the first of which is
x. First we print out the string showing the factorial. Then we print out the rest, hence the recursive call. But what should we do in the case of the empty list? Clearly we are done, so we don't need to do anything at all, so we simply
Note that this must be
return () instead of just
() because if we simply wrote
showFactorials  = () then this wouldn't be an IO action, as it needs to be. For more clarification on this, you should probably just keep reading the tutorial.