[英]Returning References from Struct Method with Lifetimes
I am trying this code:我正在尝试这个代码:
struct ByteIter<'a> {
remainder: &'a mut [u8],
index: usize,
}
impl<'a> ByteIter<'a> {
fn new(remainder: &'a mut [u8]) -> ByteIter<'a> {
ByteIter{remainder, index: 0}
}
fn next(&'a mut self) -> Option<&'a mut u8> {
if self.index >= self.remainder.len() {
None
} else {
let mut byte = &mut self.remainder[self.index];
self.index += 1;
Some(byte)
}
}
}
fn main() {
let mut a = [ 0x51, 0x52, 0x53, 0x54];
let mut bytes = ByteIter::new(&mut a);
{
let byte_1 = bytes.next();
}
let byte_2 = bytes.next();
}
Now, as I understand, the byte_1
was borrowed from bytes.现在,据我所知, byte_1
是从字节借来的。 But it's lifetime has already expired.但是它的寿命已经过了。 Still when I compile this, I see the following error:仍然当我编译这个时,我看到以下错误:
29 | let byte_2 = bytes.next();
| ^^^^^
| |
| second mutable borrow occurs here
| first borrow later used here
What is the first mutable borrow here?这里的第一个可变借用是什么? Shouldn't it be released when byte_1 goes out of scope?当 byte_1 超出范围时不应该释放它吗?
Importantly, what should I do to fix this?重要的是,我应该怎么做才能解决这个问题?
With impl<'a> ByteIter<'a>
, you declare that 'a
is the lifetime used in the structure, ie, it's the lifetime of the underlying slice.使用impl<'a> ByteIter<'a>
,您可以声明'a
是结构中使用的生命周期,即它是底层切片的生命周期。
Now, when you say fn next(&'a mut self) -> Option<&'a mut u8> {
, you're reusing the same 'a
, and you say it's the same than the returned mutable reference.现在,当您说fn next(&'a mut self) -> Option<&'a mut u8> {
,您正在重用相同的'a
,并且您说它与返回的可变引用相同。 You're saying that the returned lifetime is the same than the ByteIter
struct content.您是说返回的生命周期与ByteIter
结构内容相同。
Remove the additional lifetime constraint and the compiler will be free to compute the appropriate lifetime:删除额外的生命周期约束,编译器将可以自由计算适当的生命周期:
fn next(& mut self) -> Option<& mut u8> {
if self.index >= self.remainder.len() {
None
} else {
let mut byte = &mut self.remainder[self.index];
self.index += 1;
Some(byte)
}
}
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