相关疑难解决方法(0)

这里发生了什么(游乐场)？

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);
}

给出了这个错误:

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
   |     |     |
   |     | …

在下面的代码示例中，我试图以四种不同的方式通过a对结构X的可变引用来增加结构的成员变量。在这里，编译器为由 表示的行给出以下错误B：

error[E0502]: cannot borrow `*x` as immutable because it is also borrowed as mutable
  --> src\main.rs:17:23
   |
17 |     *x.get_a_mut() += x.get_a(); //B DOESN'T COMPILE
   |     ------------------^--------
   |     ||                |
   |     ||                immutable borrow occurs here
   |     |mutable borrow occurs here
   |     mutable borrow later used here

如果是使用可变和不可变的参照问题a在同一个表达式，为什么C和D编译？

struct X {
    a: i64,
}

impl X {
    pub fn get_a_mut(&mut self) -> &mut i64 {
        return &mut …