Haskell foldr意外地给出了错误

Question

为什么用这种方式使用foldr编写地图

map' :: (a -> b) -> [a] -> [b]
map' f xs = foldr (\x acc -> f x : acc) []

给出以下错误？

test.hs:2:13: error:
• Couldn't match expected type ‘[b]’ with actual type ‘t0 a -> [b]’
• Probable cause: ‘foldr’ is applied to too few arguments
  In the expression: foldr (\ x acc -> f x : acc) []
  In an equation for ‘map'’:
      map' f xs = foldr (\ x acc -> f x : acc) []
• Relevant bindings include
    xs :: [a] (bound at test.hs:2:8)
    f :: a -> b (bound at test.hs:2:6)
    map' :: (a -> b) -> [a] -> [b] (bound at test.hs:2:1)
|
2 | map' f xs = foldr (\x acc -> f x : acc) []
|             ^^^^^^^^^^^^^^^^^^^^^^^^^

Answer 1

你不用xs.你可以写:

map' f xs = foldr (\x acc -> f x : acc) [] xs

要么

map' f = foldr (\x acc -> f x : acc) []