Question

我想解决的问题是

Write a split(L) which returns {Even, Odd}.

我看到的可用代码

-export([even/1, odd/1, filter/2, split_using_filter/1]).

even(Integer) -> Integer rem 2 =:= 0.
odd(Integer) -> not even(Integer).

filter(F, L) -> [T || T <- L, F(T) =:= true].

split_using_filter(L) -> Even = filter(fun(X) -> X band 1 == 0 end, L),
  Odd = filter(fun(X) -> X band 1 == 1 end, L),
  {Even, Odd}.

我现在正在做的是传递

fun(X) -> X band 1 == 0 end

作为even函数，类似于odd

问题
有没有办法将even(Integer)函数作为参数传递给filter？而不是重写逻辑？感谢

Answer 1

你的代码在这里：

split_using_filter(L) -> Even = filter(fun(X) -> X band 1 == 0 end, L),
  Odd = filter(fun(X) -> X band 1 == 1 end, L),
  {Even, Odd}.

你的意思是你想要在???

之后做

split_using_filter(L) -> 
  Even = filter(fun even/1, L),
  Odd = filter(fun odd/1, L),
  {Even, Odd}.

Answer 2

Even = fun(X) -> X rem 2 =:= 0 end.
Odd  = fun(X) -> X rem 2 /= 0  end.
filter(F, L) -> [T || T <- L, F(T) =:= true].
split_using_filter(L) -> 
    {filter(Even, L), filter(Odd, L)}.

您可以阅读此http://learnyousomeerlang.com/higher-order-functions

如何在函数内传递函数作为参数？

2 个答案: