锈性状场寿命

Question

我想这是我明显遗漏的东西，但这里有..

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(())
    }
}

我读过rustlang书中关于生命的章节，但不能真正理解这里的问题所在。我要做的是构建一个基本的管道和过滤器架构。源知道它的接收器，而接收器知道它的源，因此我想存储对对象的引用。显然，这是一个终生的问题。

我首先想到的是引入生存期'a，即源/宿应该和它所链接的对象一样长。这不起作用。现在我在想，我可能需要一个比a寿命更长的'b‘，并以某种方式将其混为一谈，但正如你所看到的，这就是我困惑的地方。

Answer 1

你就快成功了：

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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