特性不能被做成一个对象 [英] The trait cannot be made into an object

问题描述

我有以下代码:

extern crate futures; // 0.1.24

use futures::Future;
use std::io;

struct Context;

pub trait MyTrait {
    fn receive(context: Context) -> Future<Item = (), Error = io::Error>;
}

pub struct MyStruct {
    my_trait: MyTrait,
}

当我尝试编译它时，我收到错误消息:

When I try to compile it I get the error message:

error[E0038]: the trait `MyTrait` cannot be made into an object
  --> src/lib.rs:13:5
   |
13 |     my_trait: MyTrait,
   |     ^^^^^^^^^^^^^^^^^ the trait `MyTrait` cannot be made into an object
   |
   = note: method `receive` has no receiver

我想我知道它为什么会发生，但是我如何从结构中引用特征?是否可以?也许还有其他方法可以实现相同的行为?

I think I know why it happens, but how do I refer to the trait from the struct? Is it possible? Maybe there are other ways to implement the same behavior?

推荐答案

您可以向结构中添加类型参数，如 Zernike 的答案，或使用特征对象.

You can either add a type parameter to your struct, as in Zernike's answer, or use a trait object.

使用类型参数对性能更有利，因为 T 的每个值都会创建一个专门的结构副本，允许静态调度.trait 对象使用动态分派，因此它允许您在运行时交换具体类型.

Using the type parameter is better for performance because each value of T will create a specialized copy of the struct, which allows for static dispatch. A trait object uses dynamic dispatch so it lets you swap the concrete type at runtime.

特征对象方法如下所示:

The trait object approach looks like this:

pub struct MyStruct<'a> {
    my_trait: &'a dyn MyTrait,
}

或者这个:

pub struct MyStruct {
    my_trait: Box<dyn MyTrait>,
}

但是，在您的情况下，MyStruct 不能成为对象，因为 receive 是一个静态方法.您需要更改它以将 &self 或 &mut self 作为它的第一个参数才能工作.还有其他限制.

However, in your case, MyStruct cannot be made into an object because receive is a static method. You'd need to change it to take &self or &mut self as its first argument for this to work. There are also other restrictions.

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