如何在Rust中编写函数？

Question

我正在尝试编写一个组成两个函数的函数。初始设计非常简单：一个函数，它接受两个函数并返回一个组合函数，然后我可以与其他函数组合，因为Rust没有rest参数。我遇到了一个由令人沮丧的无用的编译器错误构建的墙。

我的撰写功能：

fn compose<'a, A, B, C, G, F>(f: F, g: G) -> Box<Fn(A) -> C + 'a>
where
    F: 'a + Fn(A) -> B + Sized,
    G: 'a + Fn(B) -> C + Sized,
{
    Box::new(move |x| g(f(x)))
}

我想怎么用它：

fn main() {
    let addAndMultiply = compose(|x| x * 2, |x| x + 2);
    let divideAndSubtract = compose(|x| x / 2, |x| x - 2);

    let finally = compose(*addAndMultiply, *divideAndSubtract);
    println!("Result is {}", finally(10));
}

编译器不喜欢这样，无论我尝试什么，特质界限永远不会满足。错误是：

error[E0277]: the size for values of type `dyn std::ops::Fn(_) -> _` cannot be known at compilation time
  --> src/main.rs:13:19
   |
13 |     let finally = compose(*addAndMultiply, *divideAndSubtract);
   |                   ^^^^^^^ doesn't have a size known at compile-time
   |
   = help: the trait `std::marker::Sized` is not implemented for `dyn std::ops::Fn(_) -> _`
   = note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>
note: required by `compose`
  --> src/main.rs:1:1
   |
1  | / fn compose<'a, A, B, C, G, F>(f: F, g: G) -> Box<Fn(A) -> C + 'a>
2  | | where
3  | |     F: 'a + Fn(A) -> B + Sized,
4  | |     G: 'a + Fn(B) -> C + Sized,
5  | | {
6  | |     Box::new(move |x| g(f(x)))
7  | | }
   | |_^

Answer 1

作为@ljedrz points out，为了使它工作，你只需要再次引用组合函数：

let finally = compose(&*add_and_multiply, &*divide_and_subtract);

（请注意，在Rust中，约定规定变量名称应为snake_case）

---

但是，我们可以做得更好！

从Rust 1.26开始，我们可以使用abstract return types（之前的特色是作为#![feature(conservative_impl_trait)]的门控）。这可以帮助您大大简化您的示例，因为它允许您跳过生命周期，引用，Sized约束和Boxes：

fn compose<A, B, C, G, F>(f: F, g: G) -> impl Fn(A) -> C
where
    F: Fn(A) -> B,
    G: Fn(B) -> C,
{
    move |x| g(f(x))
}

fn main() {
    let add_and_multiply = compose(|x| x * 2, |x| x + 2);
    let divide_and_subtract = compose(|x| x / 2, |x| x - 2);

    let finally = compose(add_and_multiply, divide_and_subtract);
    println!("Result is {}", finally(10));
}

---

最后，既然你提到了休息参数，我怀疑你真正想要的是有一种方法可以灵活地链式组合你想要的多个函数。我为此编写了这个宏：

macro_rules! compose {
    ( $last:expr ) => { $last };
    ( $head:expr, $($tail:expr), +) => {
        compose_two($head, compose!($($tail),+))
    };
}

fn compose_two<A, B, C, G, F>(f: F, g: G) -> impl Fn(A) -> C
where
    F: Fn(A) -> B,
    G: Fn(B) -> C,
{
    move |x| g(f(x))
}

fn main() {
    let add = |x| x + 2;
    let multiply = |x| x * 2;
    let divide = |x| x / 2;
    let intermediate = compose!(add, multiply, divide);

    let subtract = |x| x - 2;
    let finally = compose!(intermediate, subtract);

    println!("Result is {}", finally(10));
}

Answer 2

只需在finally中添加引用即可：

fn main() {
    let addAndMultiply = compose(|x| x * 2, |x| x + 2);
    let divideAndSubtract = compose(|x| x / 2, |x| x - 2);

    let finally = compose(&*addAndMultiply, &*divideAndSubtract);
    println!("Result is {}", finally(10));
}

解除引用addAndMultiply或divideAndSubtract揭示了一个不是Sized的特征对象;它需要包装在Box中或引用，以便将其传递给具有Sized约束的函数。