1.What is the most general type of the function
tail
[a] -> [a]
take
Int ->[a] ->[a]
take 3 [True,True,True,True]
[Bool]
take 3
[a]->[a]
(False, ’a’)
(Bool，‘char’)
[(False，0),(True,1)]
[(Bool，char)]
([’a’, ’b’], [False, True])
( [char],[True,True])
twice f x = f (f x)
(a->b)-> a->b
add x y = x * y
Num a =>a->a->a
evens = filter even
Integral a => [a] -> [a] – [Int] -> [Int]
foldr (+) 1 [2,3,4]
10
详解：
foldr (+) 1 [2,3,4]
= 2 + (3 + (4 + 1)) – 从右向左结合
= 2 + (3 + 5)
= 2 + 8
= 10
foldr () 1 [2,3,4]
24
详解：
foldr () 1 [2,3,4]
= 2 * (3 * (4 * 1)) – 从右向左结合
= 2 * (3 * 4)
= 2 * 12
= 24
foldr (-) 1 [2,3,4]
2-(3-(4-1) = 2
详解：
foldr (-) 1 [2,3,4]
= 2 - (3 - (4 - 1)) – 从右向左结合
= 2 - (3 - 3)
= 2 - 0
= 2
foldl (-) 1 [2,3,4]
((1 - 2) - 3) - 4 = -8
详解：
foldl (-) 1 [2,3,4]
= ((1 - 2) - 3) - 4 – 从左向右结合
= (-1 - 3) - 4
= -4 - 4
= -8
sum [x | x <- [1…10], even x]
30
zip [1,2] [’a’,’b’,’c’]
[(1,‘a’),(2,‘b’)]
2.What does following expressions evaluate to？
takeWhile even [2,4,6,7,8]
takeWhile函数作用为根据一个条件从列表的开头开始，取出满足该条件的所有元素，直到遇到第一个不满足条件的元素为止。一旦遇到不满足条件的元素，它就停止，并不会继续处–理列表的剩余部分。注意其会停止
[2,4,6,8]
3.Deffne a recursive function
insert :: Int -> [Int] -> [Int]
that inserts an integer into the correct position in a sorted list of
integers.
insert :: Int -> [Int] -> [Int]
insert x [] = [x]
insert x (y:ys) = if x>y then y : insert x ys
else x : y : ys
4.Deffne the following library function using recursion:
and :: [Bool] -> Bool
and [] = True
and (x:xs) = x && and xs
5.Deffne the following library function using recursion:
reverse :: [a] -> [a]
Wrong answer：reverse (x:xs) = reverse xs : x
Correct answer：reverse (x:xs) = reverse xs ++ [x]
6.Deffne the following library function using recursion:
replicate :: Int -> a -> [a]
replicate 0 _ = []
replicate n x = x : replicate (n-1) x
7.Express the sum of the squares of the integers between 1 and 100
using a list comprehension.
sum1 0 = 0
sum1 n= n^2 + sum1(n-1)
OR
sum [ n^2 | n<-[1…100]
8.Express the sum of the products of all pairs of integers between 1
and 100 using a list comprehension.
sum [ x * y | x<- [1…100], y<-[1…100] ]
9.Given the deffnitions
(f . g) x = f (g x)
map f [ ] = [ ]
map f (x:xs) = f x : map f xs
prove the following property of map using induction (for ffnite lists):
map f (map g xs) = map (f . g) xs
见Bilibili视频.