Why is receiver declared with Receiver<Job>? - multithreading

Question

Here is the full code from the last chapter in the Rust book, under multi-threading a web-server.
use std::thread;
use std::sync::Arc;
use std::sync::Mutex;
use std::sync::mpsc;
pub struct ThreadPool {
workers: Vec<Worker>,
sender: mpsc::Sender<Job>,
}
type Job = Box<dyn FnOnce() + Send + 'static>;
impl ThreadPool {
/// Create a new ThreadPool.
///
/// The size is the number of threads in the pool.
///
/// # Panics
///
/// The `new` function will panic if the size is zero.
pub fn new(size: usize) -> ThreadPool {
assert!(size > 0);
let (sender, receiver) = mpsc::channel();
let receiver = Arc::new(Mutex::new(receiver));
let mut workers = Vec::with_capacity(size);
for id in 0..size {
workers.push(Worker::new(id, Arc::clone(&receiver)));
}
ThreadPool { workers, sender }
}
pub fn execute<F>(&self, f: F)
where
F: FnOnce() + Send + 'static,
{
let job = Box::new(f);
self.sender.send(job).unwrap();
}
}
struct Worker {
id: usize,
thread: thread::JoinHandle<()>,
}
impl Worker {
fn new(id: usize, receiver: Arc<Mutex<mpsc::Receiver<Job>>>) -> Worker {
let thread = thread::spawn(move || loop {
let job = receiver.lock().unwrap().recv().unwrap();
println!("Worker {} got a job; executing.", id);
job();
});
Worker { id, thread }
}
}
I was wondering why the new() function when implementing Worker receives an Arc<Mutex<mpsc::Receiver<Job>>> type.
I would assume it would just be Arc<Mutex<mpsc::Receiver>>. Same with sender in ThreadPool. How does mpsc::channel() know to return those types in the ThreadPool implementation?

Answer 1

Up front T in Sender<T> (and Receiver<T>) is unknown. However, when you attempt to create a ThreadPool with sender, then the compiler can infer that T must be Job. Because ThreadPool's sender field is a Sender<Job>.
Given that the mpsc::channel() returns a Sender<T> and Receiver<T> with the same T, then if Sender<T> is Sender<Job>, then Receiver<T> must also be Receiver<Job>.
This becomes apparent when you take a look at the function signature of mpsc::channel():
pub fn channel<T>() -> (Sender<T>, Receiver<T>)