How do I use static lifetimes with threads? - rust

Question

I'm currently struggling with lifetimes in Rust (1.0), especially when it comes to passing structs via channels.
How would I get this simple example to compile:
use std::sync::mpsc::{Receiver, Sender};
use std::sync::mpsc;
use std::thread::spawn;
use std::io;
use std::io::prelude::*;
struct Message<'a> {
text: &'a str,
}
fn main() {
let (tx, rx): (Sender<Message>, Receiver<Message>) = mpsc::channel();
let _handle_receive = spawn(move || {
for message in rx.iter() {
println!("{}", message.text);
}
});
let stdin = io::stdin();
for line in stdin.lock().lines() {
let message = Message {
text: &line.unwrap()[..],
};
tx.send(message).unwrap();
}
}
I get:
error[E0597]: borrowed value does not live long enough
--> src/main.rs:23:20
|
23 | text: &line.unwrap()[..],
| ^^^^^^^^^^^^^ does not live long enough
...
26 | }
| - temporary value only lives until here
|
= note: borrowed value must be valid for the static lifetime...
I can see why this is (line only lives for one iteration of for), but I can't figure out what the right way of doing this is.
Should I, as the compiler hints, try to convert the &str into &'static str?
Am I leaking memory if every line would have a 'static lifetime?
When am I supposed to use 'static anyway? Is it something I should try to avoid or is it perfectly OK?
Is there a better way of passing Strings in structs via channels?
I apologize for those naive questions. I've spent quite some time searching already, but I can't quite wrap my head around it. It's probably my dynamic language background getting in the way :)
As an aside: Is &input[..] for converting a String into a &str considered OK? It's the only stable way I could find to do this.

Answer 1

You can't convert &'a T into &'static T except by leaking memory. Luckily, this is not necessary at all. There is no reason to send borrowed pointers to the thread and keep the lines on the main thread. You don't need the lines on the main thread. Just send the lines themselves, i.e. send String.
If access from multiple threads was necessary (and you don't want to clone), use Arc<String> (in the future, Arc<str> may also work). This way the string is shared between threads, properly shared, so that it will be deallocated exactly when no thread uses it any more.
Sending non-'static references between threads is unsafe because you never know how long the other thread will keep using it, so you don't know when the borrow expires and the object can be freed. Note that scoped threads don't have this problem (which aren't in 1.0 but are being redesigned as we speak) do allow this, but regular, spawned threads do.
'static is not something you should avoid, it is perfectly fine for what it does: Denoting that a value lives for the entire duration the program is running. But if that is not what you're trying to convey, of course it is the wrong tool.

Answer 2

Think about it this way: A thread has no syntactical lifetime, i.e. the thread will not be dropped at the end of code block where it was created. Whatever data you send to the thread, you must be sure that it will live as long as the thread does, which means forever. Which means 'static.
What can go wrong in your case, is if the main loop sends a reference to a thread and destroys the string before it has been handled by the thread. The thread would access invalid memory when dealing with the string.
One option would be to put your lines into some statically allocated container but this would mean that you never can destroy those strings. Generally speaking a bad idea. Another option is to think: does the main thread actually need the line once it is read? What if the main thread transfered responsibility for line to the handling thread?
struct Message {
text: String,
}
for line in stdin.lock().lines() {
let message = Message {
text: line.unwrap(),
};
tx.send(message).unwrap();
}
Now you are transferring ownership (move) from the main thread to the handler thread. Because you move your value, no references are involved and no checks for lifetime apply anymore.