[英]Rust lifetimes, data flows into other references
我编写了以下代码,该代码过滤了可以正常工作的数据流,直到我从解析简单数字更改为也具有绑定到生存期的类型(如&str
和&[u8]
为止。
use wirefilter::{ExecutionContext, Filter, Scheme};
lazy_static::lazy_static! {
static ref SCHEME: Scheme = Scheme! {
port: Int,
name: Bytes,
};
}
#[derive(Debug)]
struct MyStruct {
port: i32,
name: String,
}
impl MyStruct {
fn scheme() -> &'static Scheme {
&SCHEME
}
fn filter_matches<'s>(&self, filter: &Filter<'s>) -> bool {
let mut ctx = ExecutionContext::new(Self::scheme());
ctx.set_field_value("port", self.port).unwrap();
ctx.set_field_value("name", self.name.as_str()).unwrap();
filter.execute(&ctx).unwrap()
}
}
fn main() -> Result<(), failure::Error> {
let data = expensive_data_iterator();
let scheme = MyStruct::scheme();
let filter = scheme
.parse("port in {2 5} && name matches \"http.*\"")?
.compile();
for my_struct in data
.filter(|my_struct| my_struct.filter_matches(&filter))
.take(2)
{
println!("{:?}", my_struct);
}
Ok(())
}
fn expensive_data_iterator() -> impl Iterator<Item = MyStruct> {
(0..).map(|port| MyStruct {
port,
name: format!("http {}", port % 2),
})
}
如果我尝试对其进行编译,则编译器将因以下原因而失败:
error[E0623]: lifetime mismatch
--> src/main.rs:26:16
|
21 | fn filter_matches<'s>(&self, filter: &Filter<'s>) -> bool {
| ----- ----------
| |
| these two types are declared with different lifetimes...
...
26 | filter.execute(&ctx).unwrap()
| ^^^^^^^ ...but data from `self` flows into `filter` here
error: aborting due to previous error
error: Could not compile `wirefilter_playground`.
To learn more, run the command again with --verbose.
Process finished with exit code 101
我首先想到的是self和filter在fn filter_matches<'s>(&self, filter: &Filter<'s>) -> bool
应该有相同的寿命,但是如果我将签名更改为fn filter_matches<'s>(&'s self, filter: &Filter<'s>) -> bool
我将开始遇到此错误:
error: borrowed data cannot be stored outside of its closure
--> src/main.rs:38:29
|
33 | let filter = scheme
| ------ ...so that variable is valid at time of its declaration
...
38 | .filter(|my_struct| my_struct.filter_matches(&filter))
| ----------- ^^^^^^^^^ -------------- cannot infer an appropriate lifetime...
| | |
| | cannot be stored outside of its closure
| borrowed data cannot outlive this closure
error: aborting due to previous error
error: Could not compile `wirefilter_playground`.
To learn more, run the command again with --verbose.
Process finished with exit code 101
我无法理解原因, Filter<'s>
绑定到延迟生成的SCHEME
,并绑定到'static
,这意味着不允许filter.execute引用&self.name.as_str()
是&self.name.as_str()
因为这样已filter.execute(&ctx)
,但不是filter.execute(&ctx)
,签名是pub fn execute(&self, ctx: &ExecutionContext<'s>) -> Result<bool, SchemeMismatchError>
应该在引用完成后立即删除引用结果没有其他生命?
为了尝试编译上面的代码,您可以使用以下Cargo.toml
:
[package]
name = "wirefilter_playground"
version = "0.1.0"
edition = "2018"
[dependencies]
wirefilter-engine = "0.6.1"
failure = "0.1.5"
lazy_static = "1.3.0"
PS:可以通过在filter_matches
方法中编译as来解决此问题,但这会很不好,因为用户在尝试过滤时只会得到parse错误,并且可能会更慢。
我看到两种解决此问题的方法:
1)延长self.name
寿命。 这可以通过将expensive_data_iterator
data_iterator收集到Vec中来实现。
--- let data = expensive_data_iterator();
+++ let data: Vec<_> = expensive_data_iterator().collect();
2)减少filter
使用寿命。
--- let filter = scheme.parse("...")?.compile();
+++ let filter = scheme.parse("...")?;
--- .filter(|my_struct| my_struct.filter_matches(&filter))
+++ .filter(|my_struct| my_struct.filter_matches(&filter.clone().compile()))
我省略了一些其他小的更改。 是的,无论哪种情况, filter_matches<'s>(&'s self, ...)
都是强制性的。
PS是的,第二个选项有效,因为my_struct
寿命超过filter
。 好吧,如果两种方法都不好,那么您可以将它们组合起来! 按块处理data
,将每个data
收集到向量中。
const N: usize = 10; // or any other size
loop {
let cur_chunk: Vec<_> = data.by_ref().take(N).collect();
if cur_chunk.is_empty() {
break;
}
let cur_filter = filter.clone().compile();
// etc
}
它仅使用O(N)内存,并且编译过滤器少N倍
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