为什么参数“借用"价值? [英] Why does the parameter "borrow" the value?

问题描述

这是一个示例:

struct X(u32);

impl X {
    fn f(&mut self, v: u32) {}
}

fn main() {
    let mut x = X(42);

    // works
    let v = x.0;
    x.f(v);

    // cannot use `x.0` because it was mutably borrowed
    x.f(x.0);
}

(Rust 游乐场)

error[E0503]: cannot use `x.0` because it was mutably borrowed
  --> src/main.rs:16:9
   |
16 |     x.f(x.0);
   |     -   ^^^ use of borrowed `x`
   |     |
   |     borrow of `x` occurs here

x.f(x.0) 不起作用的原因是什么?x.0 作为参数传递，绑定到 v 参数，类型为 u32:绝对不可能访问函数体x.0 通过参数.

What is the reason why x.f(x.0) does not work? x.0 is passed as an argument, bound to the v parameter, of type u32: there is absolutely no possibility that the function body access x.0 through the parameter.

此外，我觉得这很奇怪:

Moreover, it seems very weird to me that:

f(something);

不起作用，同时:

v = something;
f(v);

有效.

推荐答案

当你做xf(x.0)时，你借用了x来提供x 以获取对 x.0 的引用之前，>&mut self 到 f.不可能同时引用 x.0 两次.也就是说，方法 f 不能同时通过 &mut self(包括 x.0) 和对 x.0 的看似不可变的引用.

When you do x.f(x.0), you have borrowed x to provide the &mut self to f before trying to borrow x again to get a reference to x.0. It is not possible to refer to x.0 twice at the same time. That is, the method f can't have both mutable access to x via &mut self (which includes x.0) and a seemingly immutable reference to x.0 at the same time.

当使用临时变量时，你实际上得到了一个值的副本；这意味着你不再引用X中的int而是42.这是允许的.

When using a temporary variable, you actually get a copy of the value; this means you no longer refer to that int in X but 42. That's allowed.

关于非词法生命周期"评论:由于 f 使用一个很好的旧 u32 而不是对它的引用，xf(x.0) 基本上应该是等价于xf(42)，因为编译器在从x.0中取出值后可以放开x然后mut-borrowx 再次提供 &mut self 给 f.但是，编译器在编译期间很早就确定了生存期及其要​​求；因此，目前的寿命比它们必须的要宽.由于 x.0 参数，rustc 当前无法确定 x 上的借用在借用 x 之前已经结束> 用于 &mut self.正在工作解决这个问题.

Regarding the "non-lexical lifetimes" comments: Since f takes a good old u32 instead of a reference to it, x.f(x.0) should basically be equivalent to x.f(42), because the compiler can let go of x after getting the value out of x.0 and then mut-borrow x again to provide &mut self to f. However, the compiler determines the lifetimes and their requirements very early on during compilation; lifetimes are therefor currently broader than they have to be. rustc is currently unable to determine that the borrow on x due to the x.0 argument has ended before borrowing x for &mut self. There is work underway to fix this.

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