如何获得Vec元素的所有权并将其替换为其他元素? [英] How can I take ownership of a Vec element and replace it with something else?

问题描述

我正在编写以下格式的函数:

I am writing a function of the following format:

fn pop<T>(data: &mut Vec<Option<T>>) -> Option<T> {
    // Let the item be the current element at head
    let item = data[0];

    // and "remove" it.
    data[0] = None;

    item
}

当我尝试执行此操作时，我得到一个有意义的错误:

When I try to do this, I get an error which makes sense:

error[E0507]: cannot move out of index of `std::vec::Vec<std::option::Option<T>>`
 --> src/lib.rs:3:16
  |
3 |     let item = data[0];
  |                ^^^^^^^ move occurs because value has type `std::option::Option<T>`, which does not implement the `Copy` trait
  |
help: consider borrowing the `Option`'s content
  |
3 |     let item = data[0].as_ref();
  |                ^^^^^^^^^^^^^^^^
help: consider borrowing here
  |
3 |     let item = &data[0];
  |                ^^^^^^^^

当我尝试将其更改为以item为引用时，尝试将data[0]设置为None时出现错误，这也是有道理的.

When I try to change it such that item is a reference, I get an error when I try to set data[0] to None, which also makes sense.

我可以通过某种方式完成自己想做的事情吗?在我看来，无论我是否要返回引用，我都必须从Vec中获取该元素的所有权.

Is there some way I can do what I want to do? It seems to me that, whether I want to return a reference or not, I'm going to have to take ownership of the element from the Vec.

我注意到Vec有一个swap_remove方法，该方法几乎可以实现我想要的功能，只是它与Vec中已经存在的元素交换，而不是我想要的任何任意值.我知道我可以将None附加到Vec的末尾并使用swap_remove，但是我有兴趣查看是否还有另一种方法.

I noticed that Vec has a swap_remove method, which does almost exactly what I want, except that it swaps with an element already in the Vec, not with any arbitrary value as I would like. I know that I could just append None to the end of the Vec and use swap_remove, but I'm interested in seeing if there's another way.

推荐答案

使用 replace本质上是将特定位置的值替换为另一个位置，并返回先前的值.

replace essentially replaces the value in a particular location with another one and returns the previous value.

另请参阅:

• 如何在可变引用结构中为字段交换新值?

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