您如何创建Box< T>当T是特征对象时? [英] How do you create a Box<T> when T is a trait-object?

问题描述

我有以下代码

extern crate rand;
use rand::Rng;

pub struct Randomizer {
    rand: Box<Rng>,
}

impl Randomizer {
    fn new() -> Self {
        let mut r = Box::new(rand::thread_rng()); // works
        let mut cr = Randomizer { rand: r };
        cr
    }

    fn with_rng(rng: &Rng) -> Self {
        let mut r = Box::new(*rng); // doesn't work
        let mut cr = Randomizer { rand: r };
        cr
    }
}

fn main() {}

它抱怨

error[E0277]: the trait bound `rand::Rng: std::marker::Sized` is not satisfied
  --> src/main.rs:16:21
   |
16 |         let mut r = Box::new(*rng);
   |                     ^^^^^^^^ `rand::Rng` does not have a constant size known at compile-time
   |
   = help: the trait `std::marker::Sized` is not implemented for `rand::Rng`
   = note: required by `<std::boxed::Box<T>>::new`

我不明白为什么Box<T>没有在T上强加Sized的原因.

I don't understand why it requires Sized on Rng when Box<T> doesn't impose this on T.

推荐答案

有关Sized特征和绑定的更多信息-这是一个相当特殊的特征，即

More about the Sized trait and bound - it's a rather special trait, which is implicitly added to every function, which is why you don't see it listed in the prototype for Box::new:

fn new(x: T) -> Box<T>

请注意，它按值(或移动)占用x，因此您甚至需要知道调用该函数的大小.

Notice that it takes x by value (or move), so you need to know how big it is to even call the function.

相反，Box类型本身不需要 ；它使用(再次特殊的)特征绑定?Sized，表示选择退出默认的Sized绑定":

In contrast, the Box type itself does not require Sized; it uses the (again special) trait bound ?Sized, which means "opt out of the default Sized bound":

pub struct Box<T> where T: ?Sized(_);

如果您浏览一遍，可以使用一种方法来创建类型不大的Box:

If you look through, there is one way to create a Box with an unsized type:

impl<T> Box<T> where T: ?Sized
....
    unsafe fn from_raw(raw: *mut T) -> Box<T>

因此从不安全的代码中，您可以从原始指针创建一个.从那时起，所有正常的事情都开始起作用.

so from unsafe code, you can create one from a raw pointer. From then on, all the normal things work.

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