为什么Clippy建议传递Arc作为参考？

Question

I am checking Clippy findings in my code and found that the pedantic rule needless_pass_by_value might be a false positive.

It says that:

warning: this argument is passed by value, but not consumed in the function body

help: consider taking a reference instead: &Arc<Mutex<MyStruct>>

Since cloning the Arc is only reference counting, moving the Arc should not be bad idea. Does it really make any difference in terms of quality and performance to send a reference instead of a value for the Arc ?

#![warn(clippy::pedantic)]

use std::sync::{Arc, Mutex};

fn main() {
    let my_struct = MyStruct { value: 3 };
    let arc = Arc::new(Mutex::new(my_struct));

    arc_taker(arc.clone());
}

fn arc_taker(prm: Arc<Mutex<MyStruct>>) {
    prm.lock().unwrap().do_something();
}

struct MyStruct {
    value: i32,
}

impl MyStruct {
    fn do_something(&self) {
        println!("self.value: {}", self.value);
    }
}

Playground

Answer 1

Calling arc_taker(arc.clone()) increments the reference count, and returning from arc_taker decrements it again. This is useless in this case, since the arc variable of main already keeps the Arc alive during the entire call. A reference to it would already suffice. No need to bump the reference count up and down.

In your specific example, arc_taker doesn't even care that it is managed by an Arc . All it cares about is that there is a Mutex to lock , so to make your function less restrictive, just take a &Mutex<MyStruct> instead.

If you wanted to do any Arc -specific things to it, like getting the weak_count or something, taking a &Arc<..> would make sense. If your function would keep a clone of the Arc around, only then it would make sense to take an Arc by value, because then the caller can decide to give you an additional reference to it by calling .clone() (thus bumping the reference count), or to give you ownership of its own Arc (and thus not bumping the reference count).