Rust 特征字段生命周期

Rust trait field lifetime

我认为这是我明显遗漏的东西,但是这里是..

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 rustlang 书中关于生命周期的章节,但无法真正理解这里有什么问题。我想要做的是构建一个基本的管道和过滤器架构。 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。

我首先想到引入生命周期 'a 来表示 source/sink 只要它链接到的对象就会存在。这是行不通的。现在我在想我可能需要一个寿命比 'a 长的 'b 并以某种方式将其混入其中,但正如你所看到的,这就是我感到困惑的地方。

你快到了:

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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