在Rust中串联向量的最佳方法

问题描述

是否有可能在Rust中连接向量？如果是这样，是否有一种优雅的方法？我有这样的东西：

Is it even possible to concatenate vectors in Rust? If so, is there an elegant way to do so? I have something like this:

let mut a = vec![1, 2, 3];
let b = vec![4, 5, 6];

for val in &b {
    a.push(val);
}

有人知道更好的方法吗？

Does anyone know of a better way?

推荐答案

结构 std :: vec :: Vec 具有方法：

The structure std::vec::Vec has method append():

fn append(&mut self, other: &mut Vec<T>)

在您的示例中，以下代码通过变异 a 和 b ：

From your example, the following code will concatenate two vectors by mutating a and b:

fn main() {
    let mut a = vec![1, 2, 3];
    let mut b = vec![4, 5, 6];

    a.append(&mut b);

    assert_eq!(a, [1, 2, 3, 4, 5, 6]);
    assert_eq!(b, []);
}





---

或者，您可以使用将可以转换为迭代器（例如 Vec ）的所有元素附加到给定向量：

---

Alternatively, you can use Extend::extend() to append all elements of something that can be turned into an iterator (like Vec) to a given vector:

let mut a = vec![1, 2, 3];
let b = vec![4, 5, 6];

a.extend(b);
assert_eq!(a, [1, 2, 3, 4, 5, 6]);
// b is moved and can't be used anymore

请注意，向量 b 被移动而不是清空。如果您的向量包含实现，您可以将对一个向量的不可变引用传递给 extend（），以避免移动。在这种情况下，向量 b 不变：

Note that the vector b is moved instead of emptied. If your vectors contain elements that implement Copy, you can pass an immutable reference to one vector to extend() instead in order to avoid the move. In that case the vector b is not changed:

let mut a = vec![1, 2, 3];
let b = vec![4, 5, 6];

a.extend(&b);
assert_eq!(a, [1, 2, 3, 4, 5, 6]);
assert_eq!(b, [4, 5, 6]);

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