rust - 不能借用为不可变的，因为它在函数参数中也被借用为可变的

Question

这里发生了什么 ( playground )？

struct Number {
    num: i32
}

impl Number {
    fn set(&mut self, new_num: i32) {
        self.num = new_num;
    }
    fn get(&self) -> i32 {
        self.num
    }
}

fn main() {
    let mut n = Number{ num: 0 };
    n.set(n.get() + 1);
}

给出这个错误:

error[E0502]: cannot borrow `n` as immutable because it is also borrowed as mutable
  --> <anon>:17:11
   |
17 |     n.set(n.get() + 1);
   |     -     ^          - mutable borrow ends here
   |     |     |
   |     |     immutable borrow occurs here
   |     mutable borrow occurs here

但是，如果您只是将代码更改为这样，它就可以工作:

fn main() {
    let mut n = Number{ num: 0 };
    let tmp = n.get() + 1;
    n.set(tmp);
}

对我来说，它们看起来完全相同 - 我的意思是，我希望前者在编译期间转换为后者。 Rust 不会在评估下一级函数调用之前评估所有函数参数吗？

Answer 1

这一行:

n.set(n.get() + 1);

脱糖成

Number::set(&mut n, n.get() + 1);

错误信息现在可能更清楚了:

error[E0502]: cannot borrow `n` as immutable because it is also borrowed as mutable
  --> <anon>:18:25
   |
18 |     Number::set(&mut n, n.get() + 1);
   |                      -  ^          - mutable borrow ends here
   |                      |  |
   |                      |  immutable borrow occurs here
   |                      mutable borrow occurs here

当 Rust 从左到右评估参数时，该代码等同于:

let arg1 = &mut n;
let arg2 = n.get() + 1;
Number::set(arg1, arg2);

Editor's note: This code example gives an intuitive sense of the underlying problem, but isn't completely accurate. The expanded code still fails even with non-lexical lifetimes, but the original code compiles. For the full description of the problem, review the comments in the original implementation of the borrow checker

现在应该很明显出了什么问题。交换前两行可以解决这个问题，但 Rust 不会进行那种控制流分析。

这最初创建为 bug #6268 , 现在集成到RFC 2094 , non-lexical-lifetimes .如果您使用 Rust 1.36 或更新版本，NLL 会自动启用并且 your code will now compile without an error .

另见:

• Why is a mutable borrow in an argument disallowed for a mutable method call?