我有不同形状的结构:
struct Triangle { points: Vec<u8> }
struct Square { points: Vec<u8> }
struct Pentagon { points: Vec<u8> }
我有一个特质CursorReadWrite
:
use std::io::Cursor;
pub trait CursorReadWrite {
fn mwrite(&mut self, writer: &mut Cursor<Vec<u8>>) -> &mut Cursor<Vec<u8>>;
fn mread(&mut self, reader: &mut Cursor<Vec<u8>>);
}
我可以为Triangle
、Square
等实现它
impl CursorReadWrite for Triangle {
fn mwrite(&mut self, writer: &mut Cursor<Vec<u8>>) -> &mut Cursor<Vec<u8>> {
//do some work and write the data on Cursor<>
writer.write(somedata);
return writer;
}
fn mread(&mut self, reader: &mut Cursor<Vec<u8>>) {
//read data and do some work and save it in mutable self ( Triangle, Square etc)
self.points = somedata;
}
}
像这样调用函数
let csd = Cursor::new(Vec::<u8>::new());
let mut t = Triangle::default();
let new_csd = t.mwrite(&mut csd);
t.mread(&mut new_csd);
它给出了这个错误
error[E0623]: lifetime mismatch
|
25 | fn mwrite(&mut self,writer: &mut Cursor<Vec<u8>>) -> &mut Cursor<Vec<u8>>{
| -------------------- ----------------------------
| |
| this parameter and the return type are declared with different lifetimes...
...
28 | return writer;
| ^^^^^^^^^^^^ ...but data from `writer` is returned here
修复代码并不容易,因为有很多缺失的部分,但你可能希望使用显式生存期重新定义mwrite
:
pub trait CursorReadWrite<'a, 'b> {
fn mwrite(&'a mut self, writer: &'b mut Cursor<Vec<u8>>) -> &'b mut Cursor<Vec<u8>>;
fn mwread(&mut self, reader: &mut Cursor<Vec<u8>>);
}
impl<'a, 'b> CursorReadWrite<'a, 'b> for Triangle{
fn mwrite(&'a mut self, writer: &'b mut Cursor<Vec<u8>>) -> &'b mut Cursor<Vec<u8>>{
...
}
}
当您有超过 1 个输入生存期时,编译器无法判断要为输出选择哪一个。引用终身消除规则:
作为引用的每个参数都会获得其自己的生存期参数。换句话说,具有一个参数的函数获得一个生存期 参数:fn
foo<'a>(x: &'a i32)
,一个有两个参数的函数得到 两个单独的生存期参数:fn foo<'a, 'b>(x: &'a i32, y: &'b i32)
,依此类推。(...
如果有多个输入生命周期参数,但其中一个是
&self
或&mut self
因为这是一个方法,那么 self 的生命周期 分配给所有输出生存期参数。(...)