返回一个类型，而不是一个实例

Question

在Rust中有一种方法可以返回一个实现某个函数的特定类型（我不想要实例，但是类型）。像这样的东西（不以目前的形式工作）：

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

最接近的事情你问的是一个工厂函数：

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返回它返回一个对象的函数。