相关疑难解决方法(0)

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

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);
}
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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
   |     |     |
   |     | …
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rust borrow-checker

21
推荐指数
1
解决办法
3400
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borrow-checker ×1

rust ×1