file - 使用 BufReader 逐行读取文件，存储到数组中

Question

在下面的代码中创建字符串缓冲区 是我发现的最快方法，因为没有完成分配重新分配 如果我没理解错的话

pub extern fn rust_print_file() -> *mut PackChar {

    //set min size to 50 - avoid expanding when line count is 50 or less
    let mut out_vec = Vec::with_capacity(50);
    let mut curdr = env::current_dir().unwrap();//get path to file dir
    let fl_str = "file_test.txt";
    curdr.push(fl_str);//created full path to be used

    let file = BufReader::new(File::open(curdr).unwrap());

    //here i try to accommodate each line in a struct

    let mut line_index = 0;
    for line in file.lines() {

        let cur_line = line.unwrap();

        let loclbuf_size = cur_line.len();
        let mut loclbuf = String::with_capacity(buffer_size);
        //i tried two ways
        loclbuf.push_str(cur_line.unwrap()); // can't be done
        loclbuf.push_str(line.unwrap()); // can't be done too

        let pack_char = PackChar {
            int_val: line_index,
            buffer_size: loclbuf_size as i32,
            buffer: loclbuf.as_ptr() as *mut _,
        };
        line_index+=1;
        mem::forget(buffer);
        out_vec.push(pack_char);
    }
    Box::into_raw(out_vec.into_boxed_slice()) as *mut _ 
}

这是我用来传递给 C# 的结构

#[repr(C)]
pub struct PackChar {
    pub int_val: c_int,
    pub buffer: *mut c_char,
    pub buffer_size: c_int,
}

当生成一些虚拟文本时 我检查过，它正确地将数据传递给“另一端”以使用它。 但不使用读取行任务，生成如上编码的文本。

这是我尝试过的另一种方式，虽然我更喜欢上面的代码，但是这个会引发编译错误。

错误:使用移动值:buffer [E0382] on forget(buffer)

#[no_mangle]
pub extern fn rust_return_file_read_lines() -> *mut PackChar {
    let mut out_vec = Vec::with_capacity(50);

    let mut curdr = env::current_dir().unwrap();
    let fl_str = "file_test.txt";
    curdr.push(fl_str);


    let file = BufReader::new(File::open(curdr).unwrap());

    let mut lindex = 0;
    for line in file.lines() {

       let tmpbuffer = line.unwrap().into_bytes();
       let tmpbuffer_size = buffer.len();

       let pack_char = PackChar {
           int_val: lindex,
           buffer_size: tmpbuffer_size as i32,
           buffer: Box::into_raw(tmpbuffer.into_boxed_slice()) as *mut _
       };
       lindex+=1;
       mem::forget(buffer);
       out_vec.push(pack_char);

    }
    Box::into_raw(out_vec.into_boxed_slice()) as *mut _
}

编辑 只要缓冲区的类型:

缓冲区:loclbuf.as_ptr() as *mut _,

我可以将数据正确传递给 C#。 所以我怎样才能以这种方式读取行，以便每行都按照描述存储到缓冲区中？

Answer 1

现在看来，我的 visual studio 中似乎有一个错误，这不是第一次发生，但由于我是 rust 的新手，我确信代码是错误的。

这就是对我有用的，我很乐意提出意见和建议

extern crate libc;
use std::env;
use libc::c_char;
use libc::c_int;
use std::mem;
use std::io::{BufReader, BufRead};
use std::fs::File;

#[repr(C)]
pub struct PackChar {
    pub int_val: c_int,
    pub buffer: *mut c_char, // changed
    pub dbuffer_size: c_int, // added
}


#[no_mangle]
pub extern fn rust_print_file() -> *mut PackChar {
    let mut out_vec = Vec::with_capacity(50 as usize);

    let mut cwd = env::current_dir().unwrap();
    let fl_str = "file_test.txt";
    cwd.push(fl_str);
    let file = BufReader::new(File::open(cwd).unwrap());

    for (index, line) in file.lines().enumerate() {

        let buffer = line.unwrap();
        let buffer_size = buffer.len();

        let pack_char = PackChar {
            int_val: index as i32,
            dbuffer_size: buffer_size as i32,
            buffer: buffer.as_ptr() as *mut _,
        };
        mem::forget(buffer); // don't deallocate memory
        out_vec.push(pack_char);
    }

    Box::into_raw(out_vec.into_boxed_slice()) as *mut _ // changed
}