锈质性状寿命 [英] Rust trait field lifetime

问题描述

我认为这很明显我很想念，但是可以了.

I think this is something obvious I'm missing, but here goes..

use std::io;

pub trait Source<'a, T> {
    fn push(&self, t: T) -> io::Result<()>;

    fn link(&mut self, sink: &dyn Sink<'a, T>) -> io::Result<()>;
}

pub trait Sink<'a, T> {
    fn push(&self, t: T) -> io::Result<()>;

    fn link(&mut self, source: &dyn Source<T>) -> io::Result<()>;
}

pub struct SyncSource<'a, T> {
    sink: Option<&'a dyn Sink<'a, T>>,
}

impl<'a, T> SyncSource<'a, T> {
    pub fn new() -> SyncSource<'a, T> {
        SyncSource {
            sink: None,
        }
    }
}

impl<'a, T> Source<'a, T> for SyncSource<'a, T> {
    fn push(&self, t: T) -> io::Result<()> {
        match self.sink {
            Some(sink) => sink.push(t),
            None => Err(io::Error::new(io::ErrorKind::NotConnected, "no sink")),
        }
    }

    fn link(&mut self, sink: &dyn Sink<'a, T>) -> io::Result<()> {
        self.sink = Some(sink);
        Ok(())
    }
}

pub struct SyncSink<'a, T> {
    source: Option<&'a dyn Source<'a, T>>,
}

impl<'a, T> SyncSink<'a, T> {
    pub fn new() -> SyncSink<'a, T> {
        SyncSink {
            source: None,
        }
    }
}

impl<'a, T> Sink<'a, T> for SyncSink<'a, T> {
    fn push(&self, t: T) -> io::Result<()> {
        match self.source {
            Some(source) => source.push(t),
            None => Err(io::Error::new(io::ErrorKind::NotConnected, "no source")),
        }
    }

    fn link(&mut self, source: &dyn Source<T>) -> io::Result<()> {
        self.source = Some(source);
        Ok(())
    }
}

我阅读了有关终生的rustlang书，但无法真正理解这里出了什么问题.我正在尝试建立一个基本的管道和过滤器体系结构.一个源知道它的接收器，一个接收器知道它的源，因此我想存储对对象的引用.显然，这里有一个终身问题.

I read the rustlang book chapter about lifetimes but could not really understand what's wrong here. What I'm trying to do is build a basic pipe and filter architecture. A source knows its sink and a sink knows its source, thus I want to store references to the objects. Obviously, there's a lifetime issue here.

我首先想到了引入生存期'a ，以说源/接收器将一直存在，只要它链接到的对象即可.这是行不通的.现在，我想我可能需要一生的'b'，它要比'a寿命长，然后以某种方式将其扔进去，但是正如您所看到的，这就是我感到困惑的地方.

I first thought about introducing the lifetime 'a to say the source/sink shall live as long as the object it is linked to. This does not work. Now I'm thinking I may need a lifetime 'b which outlives 'a and somehow throw that in the mix, but as you can see, this is where I am confused.

推荐答案

您快到了:

use std::io;

pub trait Source<'a, T> {
    fn push(&self, t: T) -> io::Result<()>;

    // Make sure the references themselves have the 'a lifetime marker
    fn link(&'a mut self, sink: &'a dyn Sink<'a, T>) -> io::Result<()>; 
}

pub trait Sink<'a, T> {
    fn push(&self, t: T) -> io::Result<()>;

    // Make sure the references themselves have the 'a lifetime marker
    fn link(&'a mut self, source: &'a dyn Source<'a, T>) -> io::Result<()>; 
}

pub struct SyncSource<'a, T> {
    sink: Option<&'a dyn Sink<'a, T>>,
}

impl<'a, T> SyncSource<'a, T> {
    pub fn new() -> SyncSource<'a, T> {
        SyncSource {
            sink: None,
        }
    }
}

impl<'a, T> Source<'a, T> for SyncSource<'a, T> {
    fn push(&self, t: T) -> io::Result<()> {
        match self.sink {
            Some(sink) => sink.push(t),
            None => Err(io::Error::new(io::ErrorKind::NotConnected, "no sink")),
        }
    }

    // Now match the lifetime definitions that is defined in the trait
    fn link(&'a mut self, sink: &'a dyn Sink<'a, T>) -> io::Result<()> {
        self.sink = Some(sink);
        Ok(())
    }
}

pub struct SyncSink<'a, T> {
    source: Option<&'a dyn Source<'a, T>>,
}

impl<'a, T> SyncSink<'a, T> {
    pub fn new() -> SyncSink<'a, T> {
        SyncSink {
            source: None,
        }
    }
}

impl<'a, T> Sink<'a, T> for SyncSink<'a, T> {
    fn push(&self, t: T) -> io::Result<()> {
        match self.source {
            Some(source) => source.push(t),
            None => Err(io::Error::new(io::ErrorKind::NotConnected, "no source")),
        }
    }

    // Now match the lifetime definitions that is defined in the trait
    fn link(&'a mut self, source: &'a dyn Source<'a, T>) -> io::Result<()> {
        self.source = Some(source);
        Ok(())
    }
}

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