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Hands-On Data Structures and Algorithms with Rust

You're reading from   Hands-On Data Structures and Algorithms with Rust Learn programming techniques to build effective, maintainable, and readable code in Rust 2018

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Product type Paperback
Published in Jan 2019
Publisher Packt
ISBN-13 9781788995528
Length 316 pages
Edition 1st Edition
Languages
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Author (1):
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Claus Matzinger Claus Matzinger
Author Profile Icon Claus Matzinger
Claus Matzinger
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Table of Contents (15) Chapters Close

Preface 1. Hello Rust! FREE CHAPTER 2. Cargo and Crates 3. Storing Efficiently 4. Lists, Lists, and More Lists 5. Robust Trees 6. Exploring Maps and Sets 7. Collections in Rust 8. Algorithm Evaluation 9. Ordering Things 10. Finding Stuff 11. Random and Combinatorial 12. Algorithms of the Standard Library 13. Assessments 14. Other Books You May Enjoy

Moving data

The introductory snippet showed code that spawns a thread but did not pass any data into the scope. Just like any other scope, it requires either ownership of a value or at least a borrowed reference in order to work with that data. In this case, passing ownership is what we want, something that is called moving data into the scope.

If we change the snippet from the introduction to include a simple variable to print from within the thread, compilation is going to fail:

use std::thread; 

fn threading() {
let x = 10;
let handle = thread::spawn(|| {
println!("Hello from a thread, the number is {}", x);
});
handle.join().unwrap();
}

The reason for this is simple: the compiler cannot determine the lifetimes of each of the scopes (will x still be there when the thread needs it?), so it refuses to compile the code:

Compiling ch1 v0.1.0 (file:///code/ch1) 
error[E0373]: closure may outlive the current function, but it borrows `x`, which is owned by the current function
--> src/main.rs:5:32
|
5 | let handle = thread::spawn(|| {
| ^^ may outlive borrowed value `x`
6 | println!("Hello from a thread, the number is {}", x);
| - `x` is borrowed here
help: to force the closure to take ownership of `x` (and any other referenced variables), use the `move` keyword
|
5 | let handle = thread::spawn(move || {
| ^^^^^^^

As the compiler messages indicate, adding the move keyword will solve the issue! This keyword lets a thread pass ownership to a different thread; it "moves" the memory area:

fn threading() { 
let x = 10;
let handle = thread::spawn(move || {
println!("Hello from a thread, the number is {}", x);
});
handle.join().unwrap();
}

When running this snippet, the output is as follows:

Hello from a thread, the number is 10

However, for passing multiple messages into a thread or implementing an actor model, the Rust standard library offers channels. Channels are single-consumer, multi-producer queues that let the caller send messages from multiple threads.

This snippet will spawn 10 threads and have each send a number into the channel, where it will be collected into a vector after the senders have finished executing:

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

fn channels() {
const N: i32 = 10;
let (tx, rx): (Sender<i32>, Receiver<i32>) = channel();
let handles = (0..N).map(|i| {
let _tx = tx.clone();
thread::spawn(move || {
// don't use the result
let _ = _tx.send(i).unwrap();
})
});
// close all threads
for h in handles {
h.join().unwrap();
}
// receive N times
let numbers: Vec<i32> = (0..N).map(|_|
rx.recv().unwrap()
).collect();

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

As expected, the output is as follows:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

With these tools, a multithreaded application can move data between threads without the need for manual locking or the dangers of inadvertently creating side effects.

You have been reading a chapter from
Hands-On Data Structures and Algorithms with Rust
Published in: Jan 2019
Publisher: Packt
ISBN-13: 9781788995528
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