haskell - 映射严格与惰性函数

Question

(head . map f) xs = (f . head) xs

当 f 严格时，它适用于每个 xs 列表。谁能给我举个例子，为什么使用非严格的 f 它不起作用？

Answer 1

让我们采用非严格函数f = const (), 和xs = undefined。在这种情况下，我们有

map f undefined = undefined

但

f undefined = ()

所以

(head . map f) undefined = head (map f undefined) = head undefined = undefined

但

(f . head) undefined = f (head undefined) = f undefined = ()

量子点