Haskell偶数迭代

Question

I'm trying to drop all odd number from a list of integers.. however having some problems (I'm a total newbie), here's my code:

 evenfunc :: [Int] -> [Int]
 evenfunc li =
        x = head li
        y = tail li
        if even x
          then x : myevenfunc xs --take head an pass tail recursively
        else
          drop x li : myevenfunc xs --drop head from list and pass tail recursively

Its giving me a 'parse error on input '='' message when trying to run this.

What am I doing wrong here?

Answer 1

Haskell isn't an imperative language in which you order statements, but you can achieve your method by binding values to identifiers using let :

f :: Int -> Int
f x = let y = x + 5 in y * 3

But even using this your function has a few problems:

• your recursive call is wrong (use evenfunc )
• your usage of drop is probably wrong
• you never define xs anywhere

All in all, it is probably best to remodel your approach using library functions. filter does exactly what you want:

evenfunc list = filter even list

Or, even:

evenfunc = filter even

Answer 2

Of course the given solution using filter should be preferred, but the recursive version could be instructive, too:

evenfunc :: [Int] -> [Int]
evenfunc [] = []
evenfunc (x:xs) = if even x then x : tail else tail where
    tail = evenfunc xs  

The first line evenfunc [] = [] takes care of the base case - the empty list. You were missing this in your solution, and if you call head on an empty list, you get an exception.

The second line deconstructs the list using a pattern: x is the head and xs is the tail. Then we use if pretty much as in imperative languages, with the difference that it returns a value (so it is in fact closer to the ternary operator x ? y : z in Java etc). In order to avoid repetition, we define a sub-expression tail in the where clause. If it is weird to you to use things before you define them, you can use let instead, which works similar. Of course, calculating tail means to execute the recursive call.