Rust中如何为Solve trait对象指定关联类型Answer？

问题核心原因 #

你遇到的类型不匹配，本质是具体类型包装的Box（比如Box<u32>、Box<String>）与 trait 对象包装的Box<dyn Display>是完全不同的类型。即便前者实现了Display trait，Rust也不会自动将它们视为等价类型，因此返回的trait对象要求Answer为Box<dyn Display>，但你的Solver实现用的是具体类型的Box，必然会报错。

解决方案一：统一关联类型为Box<dyn Display> #

直接让所有实现Solve的结构体，将关联类型Answer统一设为Box<dyn Display>，在solve方法中返回具体值的Box时，Rust会自动将其转换为 trait 对象的Box：

use std::fmt::Display;

fn main() {
    let solver = create_solver(2);
    
    let answer = solver.solve();
    
    println!("{}", answer);
}

fn create_solver(kind: u32) -> Box<dyn Solve<Answer = Box<dyn Display>>> {
    match kind {
        1 => Box::new(SolverOne),
        2 => Box::new(SolverTwo),
        _ => unreachable!()
    }
}

trait Solve {
    type Answer: Display;
    
    fn solve(&self) -> Self::Answer;
}

struct SolverOne;

impl Solve for SolverOne {
    type Answer = Box<dyn Display>;

    fn solve(&self) -> Self::Answer {
        Box::new(3) // 自动转换为Box<dyn Display>
    }
}

struct SolverTwo;

impl Solve for SolverTwo {
    type Answer = Box<dyn Display>;
    
    fn solve(&self) -> Self::Answer {
        Box::new(String::from("three")) // 同样自动转换
    }
}

解决方案二：简化Trait（无需关联类型） #

如果不需要保留关联类型的灵活性，可以直接修改trait Solve，让solve方法直接返回Box<dyn Display>，代码会更简洁：

use std::fmt::Display;

fn main() {
    let solver = create_solver(2);
    
    let answer = solver.solve();
    
    println!("{}", answer);
}

fn create_solver(kind: u32) -> Box<dyn Solve> {
    match kind {
        1 => Box::new(SolverOne),
        2 => Box::new(SolverTwo),
        _ => unreachable!()
    }
}

trait Solve {
    fn solve(&self) -> Box<dyn Display>;
}

struct SolverOne;

impl Solve for SolverOne {
    fn solve(&self) -> Box<dyn Display> {
        Box::new(3)
    }
}

struct SolverTwo;

impl Solve for SolverTwo {
    fn solve(&self) -> Box<dyn Display> {
        Box::new(String::from("three"))
    }
}

内容的提问来源于stack exchange，提问作者stackcycle