| Safe Haskell | Safe |
|---|
S02
Exercise 3
second xs = head (tail xs)
Since tail is of type [a] -> [a], xs has to be of type
[a] and thus tail xs is of type [a]. Combining this with
the type of head (which is [a] -> a), we obtain a for
head (tail xs) and thus [a] -> a for second.
double :: Num a => a -> a Source #
double x = x*2
The operation (*) is of type Num a => a -> a -> a, hence x, needs to
be of some Num-type, resulting in Num a => a -> a for double.
palindrome :: Eq a => [a] -> Bool Source #
palindrome xs = Prelude.reverse xs == xs
reverse is of type [a] -> [a] and has no class constraints. On the contrary,
(==) is only applicable for types of the Eq class. Putting this together
results in Eq a => [a] -> Bool for palindrome.
twice :: (a -> a) -> a -> a Source #
twice f x = f (f x)
Since f is applied on the result of f x, the input as well as the
output of f need to be of the same type as x. There are no further
restrictions. Hence the type of twice is (a -> a) -> a -> a.
Exercise 4
Using the definition
filter p [] = []
filter p (x:xs)
| p x = x : filter p xs
| otherwise = filter p xs
The expression filter (== 3) [1,2,3] can be evaluated as follows (where the
justification for applying conditional equations resulting from guarded
patterns are given on the right, separated by <==):
filter (== 3) [1,2,3]
= filter (== 3) [2,3] <== 1 /= 3
= filter (== 3) [3] <== 2 /= 3
= 3 : filter (== 3) [] <== 3 == 3
= 3 : []
= [3]
Exercise 5
reverse :: [a] -> [a] Source #
For reverse we use the auxiliary function snoc ('cons' spelled
backwards), which adds an element at the end of a list. Then, using foldr, we obtain
reverse = foldr snoc []
map :: (a -> b) -> [a] -> [b] Source #
For map we use 'function composition' (.) (where (f . g) x = f (g x),
reading 'first apply g and then apply f to the result') together with
the 'cons' function (:). Using foldr, we obtain
map f = foldr ((:) . f) []