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
  --> srcmain.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 self.a;
    }

    pub fn get_a(&self) -> &i64 {
        return &self.a;
    }
}

fn my_fn(x: &mut X) {
    *x.get_a_mut() += 5; //A
    *x.get_a_mut() += x.get_a(); //B DOESN'T COMPILE
    *x.get_a_mut() += 2 * x.get_a(); //C
    *x.get_a_mut() = x.get_a() + x.get_a(); //D
}

fn main() {
    let mut x = X { a: 50 };
    my_fn(&mut x);
}

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