Haskell 列表理解：显示偶数和双奇数元素

Question

I've been working on an assignment and I just can't get it to work.

I should write a function that takes a list of numbers and gives back a list that shows even numbers but doubles all odd numbers. (Basically the same list but with doubled odds).

doubleOdd :: [Integer] -> [Integer]
doubleOdd [] = []
doubleOdd a = [x*2 | x <- a, odd x]

My problems are:

1. I'm only allowed to use +, -, *, /, ==, /=, sum, mod, elem, maximum, odd, even
2. I don't know how I get different conditions for the same value to work (like if x is even = x and in the same List Comprehension have: if x is odd = x*2)

So far I only got to print the even or the odd numbers, but never both.....

I hope someone can help me.

Answer 1

In the recursive case treat both possible cases, the case of odds and the case of evens, with the two mutually exclusive tests:

doubleOdd a = [ .... | x <- a, y <- ([x*2 | .... x] ++ [x | .... x]) ]

Since the two tests are mutually exclusive, there will be only one result.

You either must use ++ , or the if...then...else . I don't see a way to do this without one or the other.

Or maybe it can be done with some arithmetic trick but then it'd be a math question, not a Haskell question.

Answer 2

In case you are looking for a math-trick version:

> take 10 $ [ x + (x `mod` 2) * x | x <- [1..] ]
[2,2,6,4,10,6,14,8,18,10]

(you can do lot's of cheating like this with mod )