为什么我能够同时传递可变和不可变引用？

Question

根据 Rust 手册，我们不能同时使用变量作为可变和不可变引用。

但由于某种原因，当我尝试使用u32类型执行此操作时，它不会抛出任何错误。

fn main() {
    let mut num = 69;
    do_something(&mut num); // no error
    dont_do_anything(&num); // no error
    println!("{}", num);
}

fn dont_do_anything(num: &u32) {
    println!("{}", num);
}

fn do_something(num: &mut u32)  {
    *num += 1;
}

我在这里错过了什么吗？

Answer 1

你不是同时经过它们，而是一个接一个地经过它们。

fn main() {
    let mut num = 69;
    // at the following line you start to borrow num mutably
    do_something(&mut num);
    // but the mutable borrow is only used in the line above

    // at the following line you start a shared borrow of num
    dont_do_anything(&num);
    // and you only use the shared borrow in the line above.

    // then you do another shared borrow for `pritnln`
    println!("{}", num);
    // which is used just in the one line again.
}

由于任何借用之间不会发生重叠，因此不会违反任何借用规则。现在让我们看看哪些是被禁止的：

fn main() {
    let mut num = 69;
    // start mutably borrowing
    let mut_num = &mut num;
    // try to borrow immutably after we started a mutable borrow, but before we stopped using it
    dont_do_anything(&num);
    // use the mutable borrow here
    do_something(mut_num);
}

这段代码确实同时使用了可变借用和不可变借用（我们调用的行dont_do_anything），因此编译器对我们大喊大叫：

   Compiling playground v0.0.1 (/playground)
error[E0502]: cannot borrow `num` as immutable because it is also borrowed as mutable
 --> src/main.rs:6:22
  |
4 |     let mut_num = &mut num;
  |                   -------- mutable borrow occurs here
5 |     // try to borrow immutably after we started a mutable borrow, but before we stopped using it
6 |     dont_do_anything(&num);
  |                      ^^^^ immutable borrow occurs here
7 |     // use the mutable borrow here
8 |     do_something(mut_num);
  |                  ------- mutable borrow later used here