映射严格与懒惰函数

Question

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

当f严格时，它适用于每个xs列表。 任何人都可以给我例子，为什么非严格的f它不工作？

Answer 1

我们来看看非严格函数f = const()和xs = undefined。在这种情况下，我们有

map f undefined = undefined 

但

f undefined =() 

等

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

但

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

证明完毕