获取向量的最后一个元素并将其推送到同一个向量 [英] Get the last element of a vector and push it to the same vector

问题描述

我想做什么:

enum Test {
    Value1,
    Value2,
    Value3
}

fn main() {
    let mut test_vec: Vec<Test> = Vec::new();
    test_vec.push(Test::Value2);

    if let Some(last) = test_vec.last() {
        test_vec.push(*last);
    }
    //Wanted output: vector with [Test::Value2, Test::Value2]
}

我知道当我调用 last() 时，它会返回 Option<&Test>所以它会借用 test_vec 直到 if-let 块的结尾.

I understand that when I call last(), it will return Option<&Test> So it will borrow the test_vec till the end of the if-let block.

我尝试了以下方法但没有成功:

I tried the following without success:

if let Some(last) = test_vec.last().map(|v| v.clone()) {
    test_vec.push(*last);
}

//and

let last = test_vec.last().unwrap().clone();
test_vec.push(*last);

推荐答案

当试图找出借用检查器抱怨的原因时，识别所涉及的类型会很有用.

When trying to figure out why the borrow checker complains it can be useful to identify the types involved.

如果您输入:

let _: () = test_vec.last().map(|v| v.clone());

您收到一个错误，抱怨 () 和 core::option::Option<&Test> 不是同一类型.

you get an error complaining that () and core::option::Option<&Test> are not the same type.

这是怎么回事?简单地说，如果你克隆一个 &Test，你会得到一个 &Test，从而调用 .map(|v| v.clone())<Option<&Test> 上的/code> 给出了 Option<&Test>.显然，它还是借了.

What's going on? Very simply put, if you clone an &Test you get an &Test, thus calling .map(|v| v.clone()) on a Option<&Test> gives an Option<&Test>. Obviously, it still borrows.

如果您输入以下内容，则下次尝试时也会出现同样的问题:

The same problem occurs with your next attempt, if you type out:

let _: () = test_vec.last().unwrap().clone();

您收到一个错误，抱怨 () 和 &Test 不是同一类型.

you get an error complaining that () and &Test are not the same type.

通过在 Option<&Test> 上调用 unwrap，您会得到一个 &Test，然后将其克隆到 &测试.

By calling unwrap on an Option<&Test> you get an &Test which is then cloned into an &Test.

因此，问题在于缺乏取消引用.您需要提前取消引用，以避免在 Some(last):

So, the problem is a lack of dereferencing. You need to dereference earlier, to avoid borrowing test_vec in Some(last):

if let Some(last) = test_vec.last().map(|v| (*v).clone()) {
    test_vec.push(last);
}

当然，这是行不通的，因为Test 没有实现clone.一旦修复(通过#[derive(Clone)])，它就会编译.

of course, this does not work because Test does not implement clone. Once that is fixed (by #[derive(Clone)]), it compiles.

由于从引用中克隆是一种常见的需求，因此 Option(和 Iterator) 被称为 cloned:

Since cloning from references is such a common need, there is a dedicated method on Option (and Iterator) called cloned:

if let Some(last) = test_vec.last().cloned() {
    test_vec.push(last);
}

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