我不明白Haskell foldr

Question

mystery是做什麼的？

mystery xs = foldr f [] xs 

f next soFar = soFar ++ [next] 

謝謝。

Answer 1

不，foldr f [] xs =/= []。讓我們看看，

mystery xs 
    = foldr f [] xs 
    -- (assuming xs == (x:t), i.e. non-empty:) 
    = foldr f [] (x:t) 
    -- (by definition of `foldr`:) 
    = f x (foldr f [] t) 
    -- (by definition of `mystery`:) 
    = f x (mystery t) 
    = mystery t ++ [x] 

那麼，mystery做什麼？例如，它對列表[a,b,c]有什麼作用？

mystery [a,b,c] 
    = mystery [b,c] ++ [a] 
    = (mystery [c] ++ [b]) ++ [a] 
    = ((mystery [] ++ [c]) ++ [b]) ++ [a] 
    = ... 

您可以在這裏填寫圖片。