rust - 为什么 `get` 对 std::vec::Vec 和 &std::vec::Vec 都有效？

Question

这个问题已经有答案了:

What are Rust's exact auto-dereferencing rules? (4 个回答)

已关闭 3 年前。

我很困惑为什么这个函数 get适用于 Vec<T>和&Vec<T> 。我知道&Vec<T>自动转换为&[T]所以从某种意义上说，问题是为什么它适用于 Vec<T>以及 &[T] 。显然，get与&[T]一起工作，那么对于 Vec<T> 是否单独实现除了 &[T] 的实现之外？看文档，好像不是，只有一个get实现:https://doc.rust-lang.org/std/vec/struct.Vec.html#method.get

在下面的代码中，get正在作用于Vec<T> 。

fn first<T: PartialOrd + Copy>(list: Vec<T>) -> T {
    *list.get(0).unwrap()
}

fn main() {
    let number_list = vec![34, 50, 25, 100, 65];
    let result = first(number_list);
    println!("The first number is {}", result);
}

在此代码中，它作用于 &Vec<T> (又名 &[T] ):

fn first<T: PartialOrd + Copy>(list: &Vec<T>) -> T {
    *list.get(0).unwrap()
}

fn main() {
    let number_list = vec![34, 50, 25, 100, 65];
    let result = first(&number_list);
    println!("The first number is {}", result);
}

Answer 1

这是由于自动引用造成的。

来自Method Syntax在《Rust 编程语言》中...

Rust has a feature called automatic referencing and dereferencing [...] when you call a method with object.something(), Rust automatically adds in &, &mut, or * so object matches the signature of the method. In other words, the following are the same:

p1.distance(&p2);
(&p1).distance(&p2);

The first one looks much cleaner. This automatic referencing behavior works because methods have a clear receiver—the type of self. Given the receiver and name of a method, Rust can figure out definitively whether the method is reading (&self), mutating (&mut self), or consuming (self). The fact that Rust makes borrowing implicit for method receivers is a big part of making ownership ergonomic in practice.