你如何使用两个数字在Haskell中列出一个列表？

Question

Here is my problem: Declare type and define a function that takes two numbers m and n as input and returns a list containing the doubled values of all odd integers between m and n. For instance, fun 2 11 would return [6, 10, 14, 18, 22].

I don't know how I can take the two number 2 and 11 and make it into a list [2..11]. Does anyone know how to do this?

Answer 1

Use sequence generation (range syntax):

Prelude> [2 .. 11]
[2,3,4,5,6,7,8,9,10,11]

Works for symbolic values, too:

Prelude> let [m,n] = [2,11]
Prelude> [m .. n]
[2,3,4,5,6,7,8,9,10,11]

Answer 2

Didn't work with Haskell for almost two years, so correct me if I'm wrong and it doesn't work:

getDoubledOdd :: Int -> Int -> [Int]
getDoubledOdd m n = map (2*) $ filter odd [m..n]

Answer 3

A combination of list comprehension and range would be the most standard way to do it.

[ 2*x | x <- [2..11], odd x ]

The code basically says "let x loop from 2 to 11 (x <- [2..11]), and if x is odd (odd x), put 2*x into the list that will be returned".

Hope that explains.