Question

data Tree a b = Branch b (Tree a b) (Tree a b) | Leaf a
myorder :: (a -> c) -> (b -> c) -> Tree a b -> [c]
myorder p q (Leaf x) = [p x]
myorder p q (Branch x l r) = myorder p q l ++ [q x] ++ myorder p q r
tree1 = Branch "Hi"
          (Branch "All" (Leaf (1::Int)) (Leaf (2::Int)))
          (Leaf (3::Int))

问题：以下表达式的结果是什么？

> myorder id length tree1

我正在攻读考试，我想更好地了解如何在没有GHCi的情况下获得输出，因为当然我们必须手工完成，而且我真的想看看我该如何去做它。答案是[1,3,2,2,3] 如果有人能指导我，那就太棒了我也像这样画出了tree1：

     "Hi"
    /   \
 "All"   3
  / \
 1   2

Answer 1

我们需要评估

myorder id length (Branch "Hi"
                    (Branch "All" (Leaf (1::Int)) (Leaf (2::Int)))
                    (Leaf (3::Int)))

尝试按从上到下的顺序应用方程式。

myorder p q (Leaf x) = [p x]

此等式不适用，因为参数Branch "Hi" ... ...不是Leaf x形式。

myorder p q (Branch x l r) = myorder p q l ++ [q x] ++ myorder p q r

这个等式确实适用。我们有p = id，q = length，x = "Hi"，

l = Branch "All" (Leaf (1::Int)) (Leaf (2::Int)),
r = Leaf (3::Int).

因此，在右侧替换这些，表达式变为

   myorder id length (Branch "All" (Leaf (1::Int)) (Leaf (2::Int)))
++ [length "Hi"]
++ myorder id length (Leaf (3::Int)).

现在继续。

读取代码声明和定义，以便在不使用GHCi的情况下获取表达式的结果

1 个答案: