如何将实现特征的类型的迭代器的内容装箱？

Question

我正在使用某种类型的迭代器，它必须实现特征 A，并试图将其转换为具有该特征的 Boxes 的 Vec：

trait A {}

fn test2<'a, I>(iterator: I) -> Vec<Box<A + 'a>>
where
    I: IntoIterator,
    I::Item: A + 'a,
{
    iterator
        .into_iter()
        .map(|a| Box::new(a))
        .collect::<Vec<Box<A + 'a>>>()
}

然而，编译失败，说：

error[E0277]: the trait bound `std::vec::Vec<std::boxed::Box<A + 'a>>: std::iter::FromIterator<std::boxed::Box<<I as std::iter::IntoIterator>::Item>>` is not satisfied
  --> src/main.rs:11:10
   |
11 |         .collect::<Vec<Box<A + 'a>>>()
   |          ^^^^^^^ a collection of type `std::vec::Vec<std::boxed::Box<A + 'a>>` cannot be built from an iterator over elements of type `std::boxed::Box<<I as std::iter::IntoIterator>::Item>`
   |
   = help: the trait `std::iter::FromIterator<std::boxed::Box<<I as std::iter::IntoIterator>::Item>>` is not implemented for `std::vec::Vec<std::boxed::Box<A + 'a>>`
   = help: consider adding a `where std::vec::Vec<std::boxed::Box<A + 'a>>: std::iter::FromIterator<std::boxed::Box<<I as std::iter::IntoIterator>::Item>>` bound

这个错误有点道理，但后来我不明白为什么以下没有问题：

fn test<'a, T: A + 'a>(t: T) -> Box<A + 'a> {
    Box::new(t)
}

这有什么不同？我如何表达我想 Box 它们作为 A，而不是它们可能是什么类型？

Answer 1

您需要将 Box<I::Item> 转换为 Box<A>:

fn test2<'a, I>(iterator: I) -> Vec<Box<dyn A + 'a>>
where
    I: IntoIterator,
    I::Item: A + 'a,
{
    iterator
        .into_iter()
        .map(|a| Box::new(a) as Box<dyn A>)
        .collect()
}

How is [returning Box::new directly] any different?

作为:

The reason why you don't need an explicit cast in the function is that the last statement of a block is a coercion site and coercions happen implicitly at these sites. See the chapter on coercions in the nomicon for further details.

如何将实现特征的类型的迭代器的内容装箱？

How can I box the contents of an iterator of a type that implements a trait?

boxing

iterator

rust