Question

我使用此函数来过滤整数列表。我正在乞求SML，但不知道错误在哪里。

fun filter f = fn [] => []
| fn (x::xs) => if f(x)
then x::(filter f xs) else (filter f xs)

fun g(x) = if x>5 then true else false

val listTest = filter g [1, 2, 4, 6, 8, 10]

谢谢！

Answer 1

您第二个fn是多余的，请删除它：

fun filter f = fn [] => []
| (x::xs) => if f(x)
then x::(filter f xs) else (filter f xs)

然后它可以愉快地编译：

- use "a.sml";
[opening a.sml]
val filter = fn : ('a -> bool) -> 'a list -> 'a list
val g = fn : int -> bool
val listTest = [6,8,10] : int list
val it = () : unit

顺便说一句，fun g(x) = if x>5 then true else false并不是很好的风格。 fun g x = x > 5更好

Answer 2

带有fn的定义看起来像带有fun的定义，但是没有名称和箭头而不是=：

fn a0 => e0
 | a1 => e1
 | ...

这是正确的：

fun filter f = fn [] => []
                | (x::xs) => if f(x)
                             then x::(filter f xs) 
                             else (filter f xs)

但常见形式是

fun filter _ [] = []
  | filter f (x::xs) = if f x
                       then x :: filter f xs
                       else filter f xs

fun g x = x > 5

SML中的过滤器功能

2 个答案: