第二次尝试访问变量句子会导致value used here after move
,我想了解如何存储总数而不引起此问题。
我试图复制迭代器,但找不到使它起作用的方法或正确的方法。
extern crate regex;
use std::fs::File;
use std::path::Path;
use std::io::{BufReader, BufRead};
fn main() {
let sentences_path = Path::new("csv/testSentences.csv");
let sentences = BufReader::new(File::open(&sentences_path).unwrap());
let total = sentences.lines().count();
for (index, sentence) in sentences.lines().enumerate() {
let line = sentence.unwrap();
println!("Processed {} of {}, {}", index, total, line);
}
println!("Total {}", total);
}
答案 0 :(得分:3)
如果所有权先前已移动,则无法访问该值。但是,您可以使用inspect检查行,而无需更改内部行,而仅更新计数。
通过迭代线找出count
之后,您可以再次对其进行迭代,并根据需要执行逻辑。
出现此编译器错误的主要原因是:count函数消耗了您使用的行,并且您的变量已被消耗,因此无法再次访问。
这是解决方案:
use std::fs::File;
use std::io::{BufRead, BufReader, Write};
fn main() {
let path = "lines.txt";
let mut output = File::create(path).unwrap();
write!(output, "Rust\n?\nFun").unwrap();
let input = File::open(path).unwrap();
let buffered = BufReader::new(input);
let lines: Vec<_> = buffered.lines().collect();
let total = lines.len();
for (index, sentence) in lines.into_iter().enumerate() {
let line = sentence.unwrap();
println!("Processed {} of {}, {}", index, total, line);
}
println!("Total {}", total);
}
通过这种方式,您无需读取文件两次。您只需重复两次。
答案 1 :(得分:2)
不能。您需要读取文件两次。首先计算总行数。其次处理每条线。因此,您需要两个BufReader
:
extern crate regex;
use std::fs::File;
use std::io::{BufRead, BufReader};
use std::path::Path;
fn get_file_reader(path: &Path) -> impl BufRead {
BufReader::new(File::open(path).unwrap())
}
fn main() {
let sentences_path = Path::new("csv/testSentences.csv");
let sentences = get_file_reader(&sentences_path);
let total = get_file_reader(&sentences_path).lines().count();
for (index, sentence) in sentences.lines().enumerate() {
let line = sentence.unwrap();
println!("Processed {} of {}, {}", index, total, line);
}
println!("Total {}", total);
}
答案 2 :(得分:-1)
sentences.lines()
是一个迭代器。您可以借用它而不是通过调用.by_ref()
来消耗它:
let mut lines = sentences.lines();
let total = lines.by_ref().count();
for (index, sentence) in lines.enumerate() {
// ...
}