How do pipes work in Rust By Example?

Question

I'm quite confused by the output of the Rust channel head using an example :

use std::sync::mpsc::{Sender, Receiver};
use std::sync::mpsc;
use std::thread;

static NTHREADS: i32 = 3;

fn main() {
    // Channels have two endpoints: the `Sender<T>` and the `Receiver<T>`,
    // where `T` is the type of the message to be transferred
    // (type annotation is superfluous)
    let (tx, rx): (Sender<i32>, Receiver<i32>) = mpsc::channel();

    for id in 0..NTHREADS {
        // The sender endpoint can be copied
        let thread_tx = tx.clone();

        // Each thread will send its id via the channel
        thread::spawn(move || {
            // The thread takes ownership over `thread_tx`
            // Each thread queues a message in the channel
            thread_tx.send(id).unwrap();

            // Sending is a non-blocking operation, the thread will continue
            // immediately after sending its message
            println!("thread {} finished", id);
        });
    }

    // Here, all the messages are collected
    let mut ids = Vec::with_capacity(NTHREADS as usize);
    for _ in 0..NTHREADS {
        // The `recv` method picks a message from the channel
        // `recv` will block the current thread if there no messages available
        ids.push(rx.recv());
    }

    // Show the order in which the messages were sent
    println!("{:?}", ids);
}

By default, NTHREADS = 3I got the following output:

thread 2 finished
thread 1 finished
[Ok(2), Ok(1), Ok(0)]

Why is the loop println!("thread {} finished", id);in the loop forprinted in reverse order? And where is it going thread 0 finished?

When I changed to NTHREADS = 8, something more mysterious happened:

thread 6 finished
thread 7 finished
thread 8 finished
thread 9 finished
thread 5 finished
thread 4 finished
thread 3 finished
thread 2 finished
thread 1 finished
[Ok(6), Ok(7), Ok(8), Ok(9), Ok(5), Ok(4), Ok(3), Ok(2), Ok(1), Ok(0)]

The print order confused me even more, and thread 0 is always absent. How to explain this example?

I tried this on different computers and got the same results.

Answer 1

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use std::sync::mpsc::{Sender, Receiver};
use std::sync::mpsc;
use std::thread;

static NTHREADS: i32 = 3;

fn main() {
    // Channels have two endpoints: the `Sender<T>` and the `Receiver<T>`,
    // where `T` is the type of the message to be transferred
    // (type annotation is superfluous)
    let (tx, rx): (Sender<i32>, Receiver<i32>) = mpsc::channel();

    let handles: Vec<_> = (0..NTHREADS).map(|id| {
        // The sender endpoint can be copied
        let thread_tx = tx.clone();

        // Each thread will send its id via the channel
        thread::spawn(move || {
            // The thread takes ownership over `thread_tx`
            // Each thread queues a message in the channel
            thread_tx.send(id).unwrap();

            // Sending is a non-blocking operation, the thread will continue
            // immediately after sending its message
            println!("thread {} finished", id);
        })
    }).collect();

    // Here, all the messages are collected
    let mut ids = Vec::with_capacity(NTHREADS as usize);
    for _ in 0..NTHREADS {
        // The `recv` method picks a message from the channel
        // `recv` will block the current thread if there no messages available
        ids.push(rx.recv());
    }

    // Show the order in which the messages were sent
    println!("{:?}", ids);

    // Wait for threads to complete
    for handle in handles {
        handle.join().expect("Unable to join");
    }
}

, , :

thread 2 finished
thread 1 finished
[Ok(2), Ok(1), Ok(0)]
thread 0 finished

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