生锈的数组元素的并行计算 [英] Parallel computing of array elements in rust

问题描述

我是新手拉斯特（V1.0.0）和线程编程。
我尝试计算使用阵列B-数组的元素。的b阵列的每个元素可以计算独立于其他（平行）的。

I'm novice to Rust (v1.0.0) and thread-programming. I try to calculate elements of b-array using a-array. Each element of the b-array can be calculated independently of the others (parallel).

extern crate rand;
use rand::Rng;
use std::io;
use std::thread;
use std::sync::{Arc, Mutex};

fn main() {
  let mut a : [u32; 10] = [0; 10];
  let mut b = Arc::new(Mutex::new([0; 10]));
  let mut rng = rand::thread_rng();

  for x in 0..9 {
    a[x] = (rng.gen::<u32>()  % 1000) + 1;
  };

  for x in 0..4 { 
    let b = b.clone(); 
    thread::spawn(move || { let mut b = b.lock().unwrap();
      for y in 0..4 {
        b[x]   += a[y] * a[y*2+1];
        b[x+5] += a[y+1] * a[y*2];
      }
    });
  };

  thread::sleep_ms(1000); 

  for x in 0..a.len() {
    println!("a({0})={1}, b({0})={2}", x, a[x], b[x]);
  };
}

您能帮我：

1. 如果我使用：让MUT B =弧::新（互斥::新的（[U32; 10] = [0; 10]））; - >我得到错误悬而未决的名字U32。您的意思是'A'？如何设置数组元素的类型？


2. 螺纹:: sleep_ms（1000） - 它是如此粗鲁。我怎么能检查所有线程完成？


3. 我怎样才能找回我的计算B〔i]和/或在最后一节聚集线程计算的B-阵列？现在，我已经得到了错误：不能索引类型的值'的alloc ::弧::弧LT;的std ::同步互斥:: ::互斥＆LT; U32; 10＆GT;＆GT;


4. 我可以只用一个B-阵列中的存储和发送成线（使用指针）来计算B-阵列的两个元素？

1. if I use: let mut b = Arc::new(Mutex::new([u32; 10] = [0; 10])); -> I get error unresolved name 'u32'. Did you mean 'a'? How can I set the type of array element ?
2. thread::sleep_ms(1000) - It is so rudely. How can I check that all thread is finished?
3. How can I get back my calculated b[i] and/or gather thread-calculated b-arrays in the final one ? Now I've got error: cannot index a value of type 'alloc::arc::Arc<std::sync::mutex::Mutex<[u32; 10]>>'
4. Can I use only one b-array in memory and send into thread (using pointers) to calculating two elements of b-array?

感谢的解决方案。
工作code是（我已经修改了它的显示问题）：

---

Thank for solutions. Working code is (I've modified it for show problem):

extern crate rand;
use rand::Rng;
use std::thread;
use std::sync::{Arc, Mutex};

fn main() {
  let mut a : [u32; 10000] = [0; 10000];
  let b = Arc::new(Mutex::new([0u32; 10]));
  let mut rng = rand::thread_rng();

  for x in 0..10000 {
    a[x] = (rng.gen::<u32>() % 10) + 1;
  };

  for x in 0..5 {
    let b = b.clone();
    thread::spawn(move || { let mut b = b.lock().unwrap();
      println!("thread {} started", x);
      for y in 0..5000 {
        b[x]   += a[y] * a[y*2+1];
        b[x+5] += a[y+1] * a[y*2];
      };
      b[x] += a[x];
      b[x+5] -= a[x];
    println!("thread {} finished", x);
    });
  };

  thread::sleep_ms(1000);

  for x in 0..10 {
    println!("b({0})={1}", x, b.lock().unwrap()[x]);
  };
}

输出是：

thread 1 started
thread 1 finished
thread 3 started
thread 3 finished
thread 0 started
thread 0 finished
thread 2 started
thread 2 finished
thread 4 started
thread 4 finished
b(0)=149482
...
b(9)=149065

线程处理一步一步的。

Threads are processed step-by-step.

推荐答案

请注意，的clone（）在电弧法对象没有克隆的数组，只需将其增加的引用计数器的电弧。

Note that the clone() method on the Arc object does not "clone" the array, simply it increments the reference counter of the Arc.

我想你问的拉斯特并行处理数据的总体战略。您code锁定每个线程 B 的数组，所以你没有并行处理。

I think you are asking for a general strategy to process data in parallel in Rust. Your code lock the b array in each thread, so you have no parallel processing.

要你将需要整个阵列上的可变访问数组没有锁，但你不能这样做，在安全锈真正的并行处理。

To do real parallel processing you would need a mutable access to the array without a lock on the entire array but you cannot do that in safe Rust.

要做到这一点，你必须使用某种不安全的机制，这样的原始指针。

To do that you have to use some sort of unsafe mechanism, such raw pointers.

这是一个简单的例子来处理（非易变）输入向量成（可变）输出并行向量：

This is a simple example to process a (non mutable) input vector into a (mutable) output vector concurrently:

use std::thread;
use std::sync::Arc;

fn main() {
    let input = Arc::new([1u32, 2, 3, 4]);
    let output = Arc::new([0; 4]);

    let mut handles = Vec::new();

    for t in 0..4 {
        let inp = input.clone();
        let out = output.clone();
        let handle = thread::spawn(move || unsafe {
            let p = (out.as_ptr() as *mut u32).offset(t as isize);

            *p = inp[t] + (t as u32 + 1);
        });

        handles.push(handle);
    }


    for h in handles {
        h.join().unwrap();
    }

    println!("{:?}", output);
}

您仍然需要使用电弧将数据传递到线程，并有适当的生命周期管理。
那么线程里面你需要得到一个可变数据指针（ out.as_ptr（）作为* MUT U32 ），然后利用在该线程处理的项目偏移方法。

You still need to use Arc to pass data into the threads and to have a proper lifetime management. Then inside the thread you need to get a mutable pointer to the data (out.as_ptr() as *mut u32), then the item processed in that thread using the offset method.

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