锈：一生的问题，需要帮助

Question

I am making an analog of c++'s iostream for rust recently, but get confused by the lifetime system. I want to store a reference of reader or a bare reader in the IStream , so I store an Q in the structure. And than I use an phantom type D for converting Q to R .

Here's the explaination:

• R is the Reader where I actually need.
• Q is the store type of R , so Q maybe a bare R or reference &mut R .
• I use D to convert Q to R . Because borrow_from_mut(&mut R) gives me an &mut R , and borrow_from_mut(R) also gives me an &mut R . Thus it could be D: BorrowFromMut<Q> .
• and D can be converted to R by derefrence. So &mut D: DerefMut<R>
• because &mut D can be dereferenced to D , but I need the &mut D dereferenced to R , here must use a trait object to dynamic dispatch the deref_mut method, because of the absence of UFCS .(trick: let tmp: &'c mut Q = &mut *self.istream.borrow_mut(); )

Such trick makes IStream able to store both &mut R and R .

But the code can't compile because of a lifetime issue:

let tmp: &'c mut Q = &mut *self.istream.borrow_mut();
//the Q borrowed from the RefCell only valid in the block, doesn't out live 'c.

how can I solve it?

here is the code sample:

pub struct IStream<'a,'b,R:'a+'b,Q:'a+'b,Sized? D:'b> where R: Reader, D: BorrowFromMut<Q>, &'b mut D: DerefMut<R> {
    istream: Rc<RefCell<Q>>
}

impl<'a,'b,R,Q,D> Clone for IStream<'a,'b,R,Q,D> where R: Reader, D: BorrowFromMut<Q>, &'b mut D: DerefMut<R> {
    fn clone(&self) -> IStream<'a,'b,R,Q,D> {
        IStream {
            istream: self.istream.clone()
        }
    }
}

impl<'a,'b,'c,F,R,Q,D> Shr<&'b mut F,IStream<'a,'c,R,Q,D>> for IStream<'a,'c,R,Q,D> where R: Reader, F: FromStr + Default, D: BorrowFromMut<Q>, &'c mut D: DerefMut<R> {
    fn shr(mut self, output: &mut F) -> IStream<'a,'c,R,Q,D> {
        let tmp: &'c mut Q = &mut *self.istream.borrow_mut();
        let mut reader: &mut D = BorrowFromMut::borrow_from_mut(tmp);
        let mut real_reader: &DerefMut<R> = &reader;

        let mut buf = String::new(); // a string buffer

        loop {
            if let Ok(byte) = (*real_reader.deref_mut()).read_byte() {
                if byte == '\u{A}' as u8 || byte == '\u{20}' as u8 {
                    break
                } else {
                    buf.push(byte as char);
                }
            } else {
                break
            }
        }

        *output = FromStr::from_str(buf[]).unwrap_or_default();
        IStream {
            istream: self.istream.clone()
        }
    }
}

Answer 1

I think you're trying to outsmart yourself. You don't need to be able to store a reader or a mutable reference to a reader at the same time, because you can easily convert such mutable reference to a full-fledged reader. Without these complications your code will look like this (note the example of by_ref() below):

#![feature(slicing_syntax)]

use std::cell::RefCell;
use std::rc::Rc;
use std::str::FromStr;
use std::default::Default;
use std::io::ByRefReader;

pub struct IStream<R> where R: Reader {
    istream: Rc<RefCell<R>>
}

impl<R> Clone for IStream<R> where R: Reader {
    fn clone(&self) -> IStream<R> {
        IStream {
            istream: self.istream.clone()
        }
    }
}

impl<'b, F, R> Shr<&'b mut F, IStream<R>> for IStream<R> where R: Reader, F: FromStr + Default {
    fn shr(self, output: &'b mut F) -> IStream<R> {  // '
        let mut real_reader = self.istream.borrow_mut();

        let mut buf = String::new(); // a string buffer
        loop {
            if let Ok(byte) = real_reader.read_byte() {
                if byte == '\u{A}' as u8 || byte == '\u{20}' as u8 {
                    break
                } else {
                    buf.push(byte as char);
                }
            } else {
                break
            }
        }

        *output = FromStr::from_str(buf[]).unwrap_or_default();

        self.clone()
    }
}

fn main() {
    let mut stdin = std::io::stdin();
    let stdin_ref: &mut _ = &mut stdin;  // a mutable reference to a reader

    let is = IStream { istream: Rc::new(RefCell::new(stdin_ref.by_ref())) };

    let mut x: uint = 0;
    let mut y: uint = 0;

    is >> &mut x >> &mut y;

    println!("{}, {}", x, y);
}

Answer 2

Here's a reproduction of the error you mentioned:

use std::cell::RefCell;

fn foo<'a>(rc: &'a RefCell<u8>) { 
    let b: &'a u8 = &*rc.borrow();
}

fn main() {
}

This fails to compile with

borrowed value does not live long enough

By definition, the result of borrow() has a scope that is tied to the scope it's called in, you can't "trick" it by specifying a different lifetime. All that does is make the compiler tell you you can't do that.