我不明白Haskell foldr

Question

mystery做了什么？

mystery xs = foldr f [] xs

f next soFar = soFar ++ [next]

感谢。

Answer 1

不，foldr f [] xs =/= []。我们来看看，

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]

那么，mystery做了什么？例如，它对列表[a,b,c]有什么作用？

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

你可以在这里完成图片。