返回一个类型，而不是一个实例

Question

Rust 中是否有一种方法可以从函数返回实现特定特征的类型（我不想要实例，而是类型）。像这样的东西（在当前形式下不起作用）：

trait MyTrait {
    fn sum(&self, val: i64) -> i64;
}

#[derive(Debug)]
struct X {
    x: i64,
}

impl MyTrait for X {
    fn sum(&self, val: i64) -> i64 {
        self.x + 2 * val
    }
}

#[derive(Debug)]
struct Y {
    x: i64,
}

impl MyTrait for Y {
    fn sum(&self, val: i64) -> i64 {
        self.x + 3 * val
    }
}

fn from_name(name: &str) -> MyTrait {
    match name {
        "X" => X,
        "Y" => Y,
        _ => panic!("Unknown name")
    }
}

fn main() {
    let x = X{x: 21};
    let y = Y{x: 42};

    // This does not work, it is just to show the idea
    let z = from_name("X"){x: 10};

    println!("x {:?}", x.sum(3));
    println!("y {:?}", y.sum(3));
    println!("z {:?}", z.sum(3));
}

Answer 1

与您所要求的最接近的是工厂函数：

fn from_name(name: &str) -> Box<Fn(i64) -> Box<MyTrait>> {
    match name {
        "X" => Box::new(|x| Box::new(X{x: x})),
        "Y" => Box::new(|x| Box::new(Y{x: x})),
        _ => panic!("Unknown name"),
    }
}

fn example() -> i64 {
    let factory = from_name("X");
    let z = factory(10);
    z.sum(10)
}

from_name 返回一个返回对象的函数。