创建盒装特征背后的机制是如何工作的? [英] How does the mechanism behind the creation of boxed traits work?

问题描述

我无法理解盒装特征的价值是如何产生的.考虑以下代码:

trait Fooer {fn foo(&self);}i32 的 impl Fooer {fn foo(&self) { println!("Fooer on i32!");}}fn 主(){让 a = Box::new(32);//工作，创建一个 Box<i32>让 b = Box::::new(32);//工作，创建一个 Box<i32>让 c = Box::::new(32);//不起作用让 d: Box= 盒子::新(32)；//有效，创建一个 Box让 e: Box= Box::<i32>::new(32);//有效，创建一个 Box}

显然，变体 a 和 b 可以正常工作.但是，变体 c 没有，可能是因为 new 函数只接受相同类型的值，而 Fooer != i32 并非如此.变体 d 和 e 工作，这让我怀疑正在执行某种从 Box 到 Box 的自动转换.

所以我的问题是:

• 这里是否发生了某种转换?
• 如果是这样，它背后的机制是什么?它是如何工作的?(我也对底层细节感兴趣，即事物在幕后如何表示)
• 有没有办法直接从 i32 创建一个 Box?如果没有:为什么不呢?

解决方案

然而，变体 c 没有，可能是因为 new 函数只接受相同类型的值，而 Fooer != i32 并非如此.

不，这是因为没有Box没有new函数.在文档中:

<块引用>

impl;方框

pub fn new(x: T) ->方框

Box 上的大多数方法都允许 T: ?Sized，但 new 是在 impl 中定义的> 没有T: ?Sized 绑定.意味着你只能调用Box::<T>::new 当 T 是已知大小的类型时.dyn Fooer 未调整大小，因此根本没有要调用的 new 函数.

事实上，这个函数不能存在于今天的 Rust 中.Box::new 需要知道T 的具体类型，以便分配合适大小和对齐方式的内存.因此，在将它发送到 Box::new 之前，您不能擦除 T .(可以想象未来的语言扩展 可能允许函数接受未定义大小的参数；但是，尚不清楚即使 unsized_locals 是否真的会启用 Box::new 接受未定义大小的 T>.)

暂时，像 dyn Fooer 这样的 unsized 类型只能存在于胖指针"之后，即指向对象的指针和指向该对象的 Fooer 的实现.你如何得到一个胖指针?你从一个细指针开始并强制它.这就是这两行发生的事情:

let d: Box= 盒子::新(32)；//有效，创建一个 Box让 e: Box= Box::<i32>::new(32);//有效，创建一个 Box

Box::new 返回一个 Box，然后是 强制 到 Box.您可以将其视为一种转换，但是 Box 没有改变；编译器所做的只是在上面贴一个额外的指针，然后忘记它的原始类型.rodrigo 的回答更详细地介绍了这种强制转换的语言级机制.

希望所有这些都能解释为什么答案

<块引用>

有没有办法直接从 i32 创建一个 Box?

是否":i32 必须在 之前被装箱，您可以删除它的类型.这和你不能写 let x: Fooer = 10i32 的原因是一样的.

相关

• 为什么我不能写一个和 Box::new 类型相同的函数?
• 是否允许使用多态变量?
• 你如何实际使用动态大小的类型生锈?
• 为什么禁止`let ref a: Trait = Struct`?

I'm having trouble understanding how values of boxed traits come into existence. Consider the following code:

trait Fooer {
    fn foo(&self);
}

impl Fooer for i32 {
    fn foo(&self) { println!("Fooer on i32!"); }
}

fn main() {
    let a = Box::new(32);                        // works, creates a Box<i32>
    let b = Box::<i32>::new(32);                 // works, creates a Box<i32>
    let c = Box::<dyn Fooer>::new(32);           // doesn't work
    let d: Box<dyn Fooer> = Box::new(32);        // works, creates a Box<Fooer>
    let e: Box<dyn Fooer> = Box::<i32>::new(32); // works, creates a Box<Fooer>
}

Obviously, variant a and b work, trivially. However, variant c does not, probably because the new function takes only values of the same type which is not the case since Fooer != i32. Variant d and e work, which lets me suspect that some kind of automatic conversion from Box<i32> to Box<dyn Fooer> is being performed.

So my questions are:

• Does some kind of conversion happen here?
• If so, what the mechanism behind it and how does it work? (I'm also interested in the low level details, i.e. how stuff is represented under the hood)
• Is there a way to create a Box<dyn Fooer> directly from an i32? If not: why not?

解决方案

However, variant c does not, probably because the new function takes only values of the same type which is not the case since Fooer != i32.

No, it's because there is no new function for Box<dyn Fooer>. In the documentation:

impl<T> Box<T>

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

Most methods on Box<T> allow T: ?Sized, but new is defined in an impl without a T: ?Sized bound. That means you can only call Box::<T>::new when T is a type with a known size. dyn Fooer is unsized, so there simply isn't a new function to call.

In fact, that function can't exist in today's Rust. Box<T>::new needs to know the concrete type T so that it can allocate memory of the right size and alignment. Therefore, you can't erase T before you send it to Box::new. (It's conceivable that future language extensions may allow functions to accept unsized parameters; however, it's unclear whether even unsized_locals would actually enable Box<T>::new to accept unsized T.)

For the time being, unsized types like dyn Fooer can only exist behind a "fat pointer", that is, a pointer to the object and a pointer to the implementation of Fooer for that object. How do you get a fat pointer? You start with a thin pointer and coerce it. That's what's happening in these two lines:

let d: Box<Fooer> = Box::new(32);        // works, creates a Box<Fooer>
let e: Box<Fooer> = Box::<i32>::new(32); // works, creates a Box<Fooer>

Box::new returns a Box<i32>, which is then coerced to Box<Fooer>. You could consider this a conversion, but the Box isn't changed; all the compiler does is stick an extra pointer on it and forget its original type. rodrigo's answer goes into more detail about the language-level mechanics of this coercion.

Hopefully all of this goes to explain why the answer to

Is there a way to create a Box<Fooer> directly from an i32?

is "no": the i32 has to be boxed before you can erase its type. It's the same reason you can't write let x: Fooer = 10i32.

Related

• Why can't I write a function with the same type as Box::new?
• Are polymorphic variables allowed?
• How do you actually use dynamically sized types in Rust?
• Why is `let ref a: Trait = Struct` forbidden?

这篇关于创建盒装特征背后的机制是如何工作的?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！