Haskell语义调用的名称/值

Question

我是Haskell的新手，我有一个问题

h x = x : (h x)
g xs = [head xs, head xs - 1]

假设语义是按名称调用和按值调用，运行g（h 2）有什么结果？

Answer 1

让我们逐步进行。

首先，h x = x : (h x)是递归的。函数h是根据其自身定义的：给定一些x会将其放在list上，该x : x : x : ...是无限的。换句话说，由于递归从不终止，因此列表不能是有限的。但是它不会永远挂起。为什么？

因为Haskell很懒。它不会对绝对必要的表达式进行评估。在您的情况下，如果调用h 2，则根本不会在内存中创建无限列表。为什么？

因为g仅请求名为xs的给定列表的前两个元素。 Haskell足够聪明，可以只在需要的时间内扩展无限列表，而不会浪费资源。通常，这是懒惰的好处。它也可以与其他表达式一起使用，不仅限于列表。

Answer 2

“按名称呼叫”是非记忆非严格评估策略：

h x = x : (h x)
g xs = [head xs, head xs - 1]

g (h 2) = let {xs = (h 2)} in [head xs, head xs - 1]
        = [let {xs = (h 2)} in head xs, let {xs = (h 2)} in head xs - 1]
        = [head (h 2), let {xs = (h 2)} in head xs - 1]
        = [head (let {x = 2} in x : (h x)}), let {xs = (h 2)} in head xs - 1]
        = [let {x = 2} in x, let {xs = (h 2)} in head xs - 1]
        = [2, let {xs = (h 2)} in head xs - 1]
        = ....

“按需拨打电话”是“ [不受限制]”，又称“懒惰”评估策略：

h x = x : (h x)
g xs = [head xs, head xs - 1]

g (h 2) = let {xs = (h 2)} in [head xs, head xs - 1]
        = let {xs = (2 : (h 2))} in [head xs, head xs - 1]
        = let {xs = (2 : (h 2))} in [2, head xs - 1]
        = ....

“按价值致电”是
strict

评估策略：

h x = x : (h x)
g xs = [head xs, head xs - 1]

g (h 2) = let {xs = (h 2)} in [head xs, head xs - 1]
        = let {xs = (2 : (h 2))} in [head xs, head xs - 1]
        = let {xs = (2 : (2 : (h 2)))} in [head xs, head xs - 1]
        = let {xs = (2 : (2 : (2 : (h 2))))} in [head xs, head xs - 1]
        = ....

以上所有假设都在GHCi提示符下输入g (h 2)，因此需要用它完整打印。