问号在类型参数绑定中意味着什么？ [英] What does the question mark mean in a type parameter bound?

问题描述

我找到 std的定义： ：borrow :: BorrowMut ：

  pub特质借用特征<借用>：借用<借> 
其中
借入：？尺码，
 {
 fn borrow_mut（& mut self） - > & mut借用; 
 
 $ / code> 

Sized 在此类型参数中绑定的意义（借用：？大小）？

• 生锈的程式语言¹书，


• Rust Reference ，还有
  
• 什么是Sized不是实现"意思是？堆栈溢出

但没有找到解释。请在答案中提供参考。

---

¹特别参见 5.20特质

²_{和section 6.1.9 Traits}

这意味着特质是可选的。当前的语法是在 DST语法RFC 。

我知道的？的唯一特征是大小。

在这个特定的例子中，我们可以为实现 BorrowMut >未分类的类型，例如 [T] - 注意这里没有& ！

$ b pre $ impl< T> BorrowMut< [T]>对于Vec< T>


由于 Matthieu M.添加：

这是一个扩展边界的情况;在一般范围内施加约束，但在 Sized 的情况下，决定除非另有说明，通用的 T 将假定为大小。注意相反的方法是将它标记为？Sized （maybe Sized ）。

I found the definition for std::borrow::BorrowMut:

pub trait BorrowMut<Borrowed>: Borrow<Borrowed>
where
    Borrowed: ?Sized,
{
    fn borrow_mut(&mut self) -> &mut Borrowed;
}

What does the question mark in front of Sized mean in this type parameter bound (Borrowed: ?Sized)?

I consulted:

• The Rust Programming Language¹ book,
• The Rust Reference², and also
• What does "Sized is not implemented" mean? on Stack Overflow

but didn't find an explanation. Please give a reference in your answer.

---

¹ _{especially see section 5.20 Traits}
² _{and section 6.1.9 Traits}

解决方案

It means that the trait is optional. The current syntax was introduced in the DST syntax RFC.

The only trait I am aware of that works for ? is Sized.

In this specific example, we can implement BorrowMut for unsized types, like [T] — note that there's no & here!

One built-in implementation makes use of that:

impl<T> BorrowMut<[T]> for Vec<T>

As Matthieu M. adds:

This is a case of a widening bound; in general bounds impose more constraints, but in the case of Sized it was decided that unless otherwise noted a generic T would be assumed to be Sized. The way to note the opposite would be to mark it ?Sized ("maybe Sized").

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