Haskell - finding smallest Element in list

问题

I have a function that gets a List and has to return the smallest Element of it.

Unfortunately I keep getting the issue:

Parse error in pattern: minim

What could I have done wrong?

minim :: [Int] -> Int
minim []       = 0
minim [x]      = x
minim x:xs     = min x (minim xs)

min :: Int -> Int -> Int
min a b
    | a > b  = b
    | a < b  = a

回答1:

If you want to solve it the most Haskell way. I would solve it as such:

-- Does not work for empty lists (so maybe needs to be wrapped in some logic)
foldr1 min [-3,1,2,3]
-- Works for empty but needs a "default value" (in this case 0)
foldr min 0 [-3,1,2,3]

If you want to learn by implementing it yourself, then this works for me

minim :: [Int] -> Int
minim []       = 0
minim [x]      = x
minim (x:xs)   = min x (minim xs)

min :: Int -> Int -> Int
min a b
    | a > b  = b
    | a < b  = a
    | a == b = a

I would however make it a bit more safe, because is really 0 the smallest int in the list if it is empty? I think you should use Nothing as the result.

import Data.Maybe

import Prelude hiding (min)

main = print $  minim [1,3,4, 6,6,-9]

minim :: [Int] -> Maybe Int
minim []       = Nothing
minim [x]      = Just x
minim (x:xs)   = min x <$> minim xs

min :: Int -> Int -> Int
min a b
    | a > b  = b
    | a < b  = a
    | a == b = a

回答2:

You have one parameter that you want to match (the list of Ints). Where you want to match parts of that list you need to put these in brackets to show the compiler you are matching one thing. Thus the last pattern should be (x:xs).

回答3:

here's another way of implementing what you asked but without having to use auxiliary functions such as min

minElem::[Int]->Int
minElem [] = 0
minElem [x] = x
minElem (x:y:xs) 
 |x > y = minElem (y:xs)
 |x < y = minElem (x:xs)
 |x == y = minElem (x:xs)

来源：https://stackoverflow.com/questions/42530122/haskell-finding-smallest-element-in-list