问题描述
我正在学习 Rust,但遇到了一些令人困惑的行为.以下代码编译正常并按预期工作(编辑:添加了除测试函数以外的代码,以前省略):
I am learning Rust and I've run into some confusing behaviour. The following code compiles fine and works as expected (edit: added code other than test function, previously omitted):
struct Container<'a> {
contents : &'a mut i32,
}
fn main() {
let mut one = Container { contents: &mut 5 };
test(&mut one);
println!("Contents: {}",one.contents);
}
fn test<'a>(mut x : &'a mut Container) {
*x.contents += 1;
let y = x;
*y.contents += 1;
x = y;
println!("{:?}",*x.contents)
}
现在在声明中
let y = x;
类型是推断出来的.因为 x
是 &'a mut Container
类型,我认为这将是等效的:
the type is inferred. Because x
is of type &'a mut Container
, I thought that this would be equivalent:
let y: &'a mut Container = x;
但是当我这样做时,编译器会出现问题:
But when I do that, the compiler takes issue:
test_3.rs:25:5: 25:10 error: cannot assign to `x` because it is borrowed
test_3.rs:25 x = y;
^~~~~
test_3.rs:23:33: 23:34 note: borrow of `x` occurs here
test_3.rs:23 let y: &'a mut Container = x;
x
如何在正确工作的示例中没有被借用?我通过从正确工作版本中省略行 x = y;
进行测试,编译器说:
How is x
not borrowed by that point in the correctly working example? I tested by omitting the line x = y;
from the correctly working version and the compiler said:
test_3.rs:24:13: 24:14 note: `x` moved here because it has type `&mut Container<'_>`, which is moved by default
因此,当我没有明确定义类型而是借用其他方式时,我正在采取行动.发生了什么,我如何在明确给出类型的同时获得与以前相同的行为,以及是什么导致在一种情况下移动行为但在另一种情况下借用?
So I'm getting a move when I don't explicitly define the type but a borrow otherwise. What is going on, how do I get the same behavior as before while explicitly giving the type, and what is causing move behavior in one case but borrow in the other?
用完整程序编辑
推荐答案
什么时候做
let y = x;
移动发生了.x
被清空,可以这么说,所有权转移到 y
.
a move happens. x
is emptied, so to speak, and ownership is transferred to y
.
当你做任何一个
let y: &mut _ = x;
let y: &'a mut _ = x;
x
被重新借用以帮助匹配生命周期.这大致翻译为
x
is reborrowed to aid matching the lifetimes. This roughly translates to
let y: &mut _ = &mut *x;
let y: &'a mut _ = &mut *x;
这使得 x
非空,持有别名可变借用.因此分配给它必须等待 y
被销毁.或者,您可以预先移动它
This leaves x
non-empty, holding an aliased mutable borrow. Assigning to it thus must wait for y
to be destroyed. Alternatively, you can pre-move it
let tmp = x;
let y: &'a mut _ = tmp;
我承认这是不明显的行为,很遗憾你不能在不借用整个值的情况下借用值的内容.
I'll admit it's nonobvious behaviour, and it's a shame that you can't borrow the contents of a value without borrowing the whole value.
这篇关于类型推断和借用与所有权转让的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!