Question

我正在尝试迭代&[u8]的可变缓冲区，只要有--删除所有字节，直到遇到'\n'

我尝试了这种方法，它给出了一个我无法解决的错误

fn remove_comments_in_place<'a>(buffer: &'a mut [u8]) {
    #[derive(PartialEq)]
    enum Mode {
        COMMENT,
        CODE,
    }
    let mut mode = Mode::CODE;
    let mut iter = buffer.iter_mut().peekable();
    while let Some(ch) = iter.next() {
        match ch {
            // Look 2 chars ahead to identify comments
            &mut b'-' => {
                if let Some(&&mut ref mut hyphen) = iter.peek() {
                    if hyphen == &mut b'-' {
                        *hyphen = b' ';
                        *ch = b' ';
                        mode = Mode::COMMENT;
                    }
                }
            }
            &mut b'\r' => *ch = b' ',
            &mut b'\n' => {
                *ch = b' ';
                if mode == Mode::COMMENT {
                    mode = Mode::CODE;
                }
            }
            _ => if mode == Mode::COMMENT {
                *ch = b' '
            },
        }
    }
}

我得到的错误是：

error[E0389]: cannot borrow data mutably in a `&` reference
  --> src/main.rs:13:35
   |
13 |                 if let Some(&&mut ref mut hyphen) = iter.peek() {
   |                                   ^^^^^^^^^^^^^^ assignment into an immutable reference

我也尝试将作业更改为

&mut b'-' => {
    if let Some(&&mut hyphen) = iter.peek() {
        if hyphen == b'-' {
            hyphen = b' ';
            *ch = b' ';
            mode = Mode::COMMENT;
        }
    }
}

但收到以下错误：

warning: value assigned to `hyphen` is never read
  --> src/main.rs:15:25
   |
15 |                         hyphen = b' ';
   |                         ^^^^^^
   |
   = note: #[warn(unused_assignments)] on by default

error[E0384]: re-assignment of immutable variable `hyphen`
  --> src/main.rs:15:25
   |
13 |                 if let Some(&&mut hyphen) = iter.peek() {
   |                                   ------ first assignment to `hyphen`
14 |                     if hyphen == b'-' {
15 |                         hyphen = b' ';
   |                         ^^^^^^^^^^^^^ re-assignment of immutable variable

This question suggests my destructuring is correct，那么我如何使用hyphen分配到Option<&mut T>？

Answer 1

你可能应该回过头来自我介绍mutability with the book。 Peekable::peek返回不可变引用：

library(dplyr)
library(tidyr)
inputTest %>%
       tbl_df %>% 
       gather(date, value, matches("^\\d+") )

因为它是不可变的你......不能改变它。

impl<I> Peekable<I>
where
    I: Iterator, 
{
    fn peek(&mut self) -> Option<&<I as Iterator>::Item>
    //                           ^^^^^^^^^^^^^^^^^^^^^^
}

解构一个可变的peekable迭代器

1 个答案: