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

这是怎么回事 (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);
}

出现此错误：

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，如果您只需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, now it is integrated into RFC 2094, non-lexical-lifetimes. If you use Rust 1.36 or newer, NLL is enabled automatically and your code will now compile without an error。

另请参阅：

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

Cannot borrow as immutable because it is also borrowed as mutable in function arguments

rust

borrow-checker