rust - 嵌套结构：“借来的值不够长”

问题描述

我一直在努力创建一个Table可以解析为Val特征对象以及其他数据类型（Number作为示例实现）的结构，并且可以在 Rust Playground 中看到。我无法解决这个借用/终身问题，但我的想法是，Number两者Table都是Val对象。但是，任何将对象写入String缓冲区的尝试都会导致编译失败。

我在这里做错了什么？借用检查器以这种方式运行的基本原理是什么？

use std::fmt;
use std::fmt::Write;

struct Number(f64);

struct Table<'a> {
    rows: Vec<Row<'a>>
    // Will have column vector as a member
}

impl <'a>Table<'a> {
    fn new(rows: Vec<Vec<Box<dyn Val>>>) -> Table {
        let mut table = Table {
            rows: rows.into_iter().map(|r| {
                Row {
                    parent: std::ptr::null_mut(),
                    cells: r
                }
            }).collect()
        };

        let parent = &mut table as *mut Table;

        table.rows = table.rows.into_iter().map(|mut r| {
            r.parent = parent;
            r
        }).collect();

        table
    }
}

struct Row<'a> {
    parent: *mut Table<'a>,
    cells: Vec<Box<dyn Val<'a>>>
}

impl <'a>Row<'a> {
    fn to_str(&'a self, buf: &mut String) -> fmt::Result {
        let mut cell_iter = self.cells.iter().enumerate().peekable();
        let _parent = unsafe { self.parent.as_ref() }; // Real implementation will need parent ref
        while let Some((idx,_c)) = cell_iter.next() { // Real implementation will cycle through columns in parent struct
            match self.cells.get(idx) {
                Some(v) => v.to_str(buf),
                None => Ok(())
            }?;
            if let Some(_) = cell_iter.peek() {
                write!(buf, ",")?;
            }
        }
        Ok(())
    }
}

pub trait Val<'a> {
    fn to_str(&'a self, buf: &mut String) -> fmt::Result;
}

pub trait ObjWriter<'a> {
    fn to_str(&'a self, buf: &'a mut String) -> fmt::Result;
}

impl <'a>ObjWriter<'a> for dyn Val<'a> {
    fn to_str(&'a self, buf: &mut String) -> fmt::Result { self.to_str(buf) }
}

impl <'a>Val<'a> for Table<'a> {
    fn to_str(&'a self, buf: &mut String) -> fmt::Result {
        write!(buf, "(START TABLE:")?;
        let mut row_iter = self.rows.iter().peekable();
        while let Some(r) = row_iter.next() {
            r.to_str(buf)?;
            write!(buf, "\n")?;
        }
        write!(buf, "END TABLE)")
    }
}

impl Number {
    fn to_str(&self, buf: &mut String) -> fmt::Result {
        write!(buf,"{}",self.0)
    }
}

impl <'a>Val<'a> for Number {
    fn to_str(&self, buf: &mut String) -> fmt::Result {
        self.to_str(buf)
    }
}

fn main() {
    let table = Table::new(vec![
        vec![Box::new(Number(0.5)),Box::new(Table::new(Vec::new()))],
        vec![Box::new(Table::new(Vec::new())),Box::new(Number(0.5))],
    ]);

    let mut buf = String::new();
    table.to_str(&mut buf);
    println!("{}",buf)
}

error[E0597]: `table` does not live long enough
   --> src/main.rs:98:5
    |
92  |       let table = Table::new(vec![
    |  ____________________________-
93  | |         vec![Box::new(Number(0.5)),Box::new(Table::new(Vec::new()))],
94  | |         vec![Box::new(Table::new(Vec::new())),Box::new(Number(0.5))],
95  | |     ]);
    | |_____- cast requires that `table` is borrowed for `'static`
...
98  |       table.to_str(&mut buf);
    |       ^^^^^ borrowed value does not live long enough
99  |       println!("{}",buf)
100 |   }
    |   - `table` dropped here while still borrowed

标签： rustlifetimeborrow-checkerborrowing

rust - 嵌套结构：“借来的值不够长”

问题描述

解决方案

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