Question

我是Rust的新手，我正在努力解决借贷问题。

我想加载一个Vec<Vec<f64>>矩阵，然后并行处理它。但是当我尝试编译这段代码时，我会在error: capture of moved value: `matrix` [E0382]行获得let _ =。

这个矩阵应该是线程的只读，他们不会修改它。我怎样才能通过它只读它并使'＃34;移动的值＆＃34;错误消失了？

fn process(matrix: &Vec<Vec<f64>>) {
    // do nothing for now
}

fn test() {
    let filename = "matrix.tsv";
    // loads matrix into a Vec<Vec<f64>>
    let mut matrix = load_matrix(filename);

    // Determine number of cpus
    let ncpus = num_cpus::get();
    println!("Number of cpus on this machine: {}", ncpus);
    for i in 0..ncpus {
        // In the next line the "error: capture of moved value: matrix" happens
        let _ = thread::spawn(move || {
            println!("Thread number: {}", i);
            process(&matrix);
        });
    }
    let d = Duration::from_millis(1000 * 1000);
    thread::sleep(d);
}

Answer 1

将要共享的对象包装到Arc中，表示原子引用计数（指针）。对于每个线程，克隆此指针并将克隆的所有权传递给线程。当被包装的对象不再被任何东西使用时，它将被解除分配。

fn process(matrix: &Vec<Vec<f64>>) {
    // do nothing for now
}

fn test() {
    use std::sync::Arc;
    let filename = "matrix.tsv";
    // loads matrix into a Vec<Vec<f64>>
    let mut matrix = Arc::new(load_matrix(filename));

    // Determine number of cpus
    let ncpus = num_cpus::get();
    println!("Number of cpus on this machine: {}", ncpus);
    for i in 0..ncpus {
        let matrix = matrix.clone();
        let _ = thread::spawn(move || {
            println!("Thread number: {}", i);
            process(&matrix);
        });
    }
    let d = Duration::from_millis(1000 * 1000);
    thread::sleep(d);
}

将矩阵（Vec <vec <f64>＆gt;）只读传递给多个线程

1 个答案: