Easy Rust

Channels

A channel is an easy way to use many threads that send to one place. They are fairly popular because they are pretty simple to put together. You can create a channel in Rust with std::sync::mpsc. mpsc means "multiple producer, single consumer", so "many threads sending to one place". To start a channel, you use channel(). This creates a Sender and a Receiver that are tied together. You can see this in the function signature:

#![allow(unused)]
fn main() {
// 🚧
pub fn channel<T>() -> (Sender<T>, Receiver<T>)
}

So you have to choose one name for the sender and one for the receiver. Usually you see something like let (sender, receiver) = channel(); to start. Because it's generic, Rust won't know the type if that is all you write:

use std::sync::mpsc::channel;

fn main() {
    let (sender, receiver) = channel(); // ⚠️
}

The compiler says:

error[E0282]: type annotations needed for `(std::sync::mpsc::Sender<T>, std::sync::mpsc::Receiver<T>)`
  --> src\main.rs:30:30
   |
30 |     let (sender, receiver) = channel();
   |         ------------------   ^^^^^^^ cannot infer type for type parameter `T` declared on the function `channel`
   |         |
   |         consider giving this pattern the explicit type `(std::sync::mpsc::Sender<T>, std::sync::mpsc::Receiver<T>)`, where
the type parameter `T` is specified

It suggests adding a type for the Sender and Receiver. You can do that if you want:

use std::sync::mpsc::{channel, Sender, Receiver}; // Added Sender and Receiver here

fn main() {
    let (sender, receiver): (Sender<i32>, Receiver<i32>) = channel();
}

but you don't have to. Once you start using the Sender and Receiver, Rust can guess the type.

So let's look at the simplest way to use a channel.

use std::sync::mpsc::channel;

fn main() {
    let (sender, receiver) = channel();

    sender.send(5);
    receiver.recv(); // recv = receive, not "rec v"
}

Now the compiler knows the type. sender is a Result<(), SendError<i32>> and receiver is a Result<i32, RecvError>. So you can use .unwrap() to see if the sending works, or use better error handling. Let's add .unwrap() and also println! to see what we get:

use std::sync::mpsc::channel;

fn main() {
    let (sender, receiver) = channel();

    sender.send(5).unwrap();
    println!("{}", receiver.recv().unwrap());
}

This prints 5.

A channel is like an Arc because you can clone it and send the clones into other threads. Let's make two threads and send values to receiver. This code will work, but it is not exactly what we want.

use std::sync::mpsc::channel;

fn main() {
    let (sender, receiver) = channel();
    let sender_clone = sender.clone();

    std::thread::spawn(move|| { // move sender in
        sender.send("Send a &str this time").unwrap();
    });

    std::thread::spawn(move|| { // move sender_clone in
        sender_clone.send("And here is another &str").unwrap();
    });

    println!("{}", receiver.recv().unwrap());
}

The two threads start sending, and then we println!. It might say Send a &str this time or And here is another &str, depending on which thread finished first. Let's make a join handle to make them wait.

use std::sync::mpsc::channel;

fn main() {
    let (sender, receiver) = channel();
    let sender_clone = sender.clone();
    let mut handle_vec = vec![]; // Put our handles in here

    handle_vec.push(std::thread::spawn(move|| {  // push this into the vec
        sender.send("Send a &str this time").unwrap();
    }));

    handle_vec.push(std::thread::spawn(move|| {  // and push this into the vec
        sender_clone.send("And here is another &str").unwrap();
    }));

    for _ in handle_vec { // now handle_vec has 2 items. Let's print them
        println!("{:?}", receiver.recv().unwrap());
    }
}

This prints:

"Send a &str this time"
"And here is another &str"

Now let's make a results_vec instead of printing.

use std::sync::mpsc::channel;

fn main() {
    let (sender, receiver) = channel();
    let sender_clone = sender.clone();
    let mut handle_vec = vec![];
    let mut results_vec = vec![];

    handle_vec.push(std::thread::spawn(move|| {
        sender.send("Send a &str this time").unwrap();
    }));

    handle_vec.push(std::thread::spawn(move|| {
        sender_clone.send("And here is another &str").unwrap();
    }));

    for _ in handle_vec {
        results_vec.push(receiver.recv().unwrap());
    }

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

Now the results are in our vec: ["Send a &str this time", "And here is another &str"].

Now let's pretend that we have a lot of work to do, and want to use threads. We have a big vec with 1000 items, all 0. We want to change each 0 to a 1. We will use ten threads, and each thread will do one tenth of the work. We will create a new vec and use .extend() to put the work in.

use std::sync::mpsc::channel;

fn main() {
    let (sender, receiver) = channel();
    let hugevec = vec![0; 1000];
    let mut newvec = vec![];

    for i in 0..10 {
        let sender_clone = sender.clone();
        let mut work: Vec<u8> = Vec::with_capacity(hugevec.len() / 10); // new vec to put the work in. 1/10th the size
        work.extend(&hugevec[i*100..(i+1)*100]); // first part gets 0..100, next gets 100..200, etc.
        let handle = std::thread::spawn(move || { // make a handle

            for number in work.iter_mut() { // do the actual work
                *number += 1;
            };
            sender_clone.send(work).unwrap(); // use the sender_clone to send the work to the receiver
        });

        handle.join().unwrap(); // stop the thread until it's done
        newvec.push(receiver.recv().unwrap()); // push the results from receiver.recv() into the vec
    }

    // Now we have a Vec<Vec<u8>>. To put it together we can use .flatten()
    let newvec = newvec.into_iter().flatten().collect::<Vec<u8>>(); // Now it's one vec of 1000 u8 numbers
}

If you print this you can see 1000 number 1s.