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。 任何人都可以举例说明，为什么非严格的f它不起作用？

Answer 1

我们选择非严格函数f = const ()和xs = undefined。在这种情况下，我们有

map f undefined = undefined

但

f undefined = ()

等等

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

但

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

Q.E.D。