如果 x 在列表中，则返回 true，否则返回 False（递归）

Question

I'm writing a short recursive function to take a list as input and output a Bool. (I'm haskell beginner) So far I can detect if the first element is a 3 or not, but I'm not sure how to use recursion to check the rest of the list.

func :: [Int] -> Bool
func [] = False
func (x:xs)
  | (x == 3)           = True
  | otherwise          = False

Answer 1

I'm new to Haskell too.

by a little change to your code, it could be re-written as

func :: [Int] -> Bool
func [] = False
func (x:xs)
  | x == 3    = True
  | otherwise = func xs

explain:

• if list is empty: there is no 3
• if list is no empty:
  1. if head is 3, then we have 3
  2. otherwise we should check rest of the list, so answer of "3 is in list" is equivalent to " x is in xs".

if you accept a little change, i can suggest implementing with OR (and help of lazy evaluation).

func :: [Int] -> Bool
func [] = False
func (x:xs) = x==3 || func xs

it is really same as the upper code, but with less lines.

• if head is 3, return True.
• if head is not 3, check rest of list.

at last, I want to introduce you elem function, it works as: get an element and a list, return True if a is in list, otherwise False. It is exactly what we want here, so i write:

containsThree :: [Int] -> Bool
containsThree = elem 3  

also note that I used point-free style, if you are not familiar, second line is same as:

containsThree xs = elem 3 xs  

Good luck learning Haskell.