Question

我与借阅检查员有争执。我的问题有点复杂，但对于这种情况，我使用缓冲区结构。我的缓冲区有一个函数safe_write_to_slot，它首先检索第一个空元素（返回结果为Ok（位置）或Err（错误消息）），然后将值写入该检索到的位置。然而问题是，当我将检索到的位置分配给值时，会抱怨我在几行之后再次重复使用self。我如何首先调用一个返回结果的（self）函数，然后继续自己做一些动作？

use std::result::Result;

struct Elems {
    pub elems : Vec<int>,
}

impl Elems {
    pub fn new() -> Elems {
        Elems{elems: vec![0,0,0,0,0,0]}
    }

    pub fn safe_write_to_slot(&mut self, elem : uint) -> Result<(), &str> {
        let loc = try!(self.get_slot());

        self.unsafe_write_to_slot(loc);

        Ok(())
    }

    pub fn get_slot(&self) -> Result<uint, &str>{
        let mut loc = -1i;

        for x in range(0, self.elems.len()) {
            if *self.elems.get(x) == 0 {
                loc = x as int;
            }
        }

        if loc != -1 { Ok(loc as uint) } else { Err("No free slots") }
    }

    fn unsafe_write_to_slot(&mut self, elem : uint) {
        self.elems[elem] = 1;
    }
}

我得到的错误是：

   Compiling borrow v0.0.1 (file:///borrow)
main.rs:19:9: 19:13 error: cannot borrow `*self` as mutable because it is also borrowed as immutable
main.rs:19         self.unsafe_write_to_slot(loc);
                   ^~~~
main.rs:17:24: 17:28 note: previous borrow of `*self` occurs here; the immutable borrow prevents subsequent moves or mutable borrows of `*self` until the borrow ends
main.rs:17         let loc = try!(self.get_slot());
                                  ^~~~
/main.rs:17:19: 17:41 note: expansion site
main.rs:22:6: 22:6 note: previous borrow ends here
main.rs:16     pub fn safe_write_to_slot(&mut self, elem : uint) -> Result<(), &str> {
/main.rs:22     }
               ^
main.rs:37:9: 37:29 error: cannot assign to immutable dereference (dereference is implicit, due to indexing)
main.rs:37         self.elems[elem] = 1;
                   ^~~~~~~~~~~~~~~~~~~~
error: aborting due to 2 previous errors
Could not compile `borrow`.

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Answer 1

终身推断导致了这个问题。

get_slot方法被解释为：

pub fn get_slot<'a>(&'a self) -> Result<uint, &'a str> {

结果绑定到与self相同的生命周期，这会导致self保持冻结状态，直到结果被删除。但是，您不希望将self的生命周期与&str相关联，因为您只返回字符串文字。通过将&str更改为&'static str和get_slot中的safe_write_to_slot，您将不会再收到错误，因为self在调用时不会被视为借用{ {1}}。

在同一功能中多次使用self

1 个答案: