46

There's a lot of Rust documentation about using modules, but I haven't found an example of a Cargo binary that has multiple modules, with one module using another. My example has three files inside the src folder. Modules a and b are at the same level. One is not a submodule of another.

main.rs:

mod a;

fn main() {
    println!("Hello, world!");
    a::a();
}

a.rs:

pub fn a() {
    println!("A");
    b::b();
}

and b.rs:

pub fn b() {
    println!("B");
}

I've tried variations of use b and mod b inside a.rs, but I cannot get this code to compile. If I try to use use b, for example, I get the following error:

 --> src/a.rs:1:5
  |
1 | use b;
  |     ^ no `b` in the root. Did you mean to use `a`?

What's the right way to have Rust recognize that I want to use module b from module a inside a cargo app?

Improve this question
1
  • 1
    Interesting, I thought I was the only one who did not understand that part in the docs. However, it was easy to know by experimenting with it. Commented Jan 3, 2018 at 14:26

3 Answers 3

Reset to default
52

You'll have to include b.rs somewhere, typically with mod b;. If b is a child of a (instead of being a sibling of a), there are two ways to do this:

  • Recommended: rename a.rs into a/mod.rs and b.rs into a/b.rs. Then you can mod b; in a/mod.rs.
  • Instead, you can just #[path = "b.rs"] mod b; in a.rs without renaming sources.

If b is intended to be a sibling of a (instead of being a child of a), you can just mod b; in main.rs and then use crate::b; in a.rs.

Improve this answer
Sign up to request clarification or add additional context in comments.

3 Comments

Thanks. That seems counterintuitive. I need to mod b in main.rs in order to use b in a.rs.
The idea is that mod is for building the module tree and use is for short-circuiting it. The absolute path ::b::b(); works without use b;. (Note that the syntax for absolute paths will change in future versions of Rust, in favor of RFC 2126.)
@user8360684 Agreed
26

The method from the accepted answer doesn't work for me in Rust 1.33. Instead, I use the sibling module like this:

use crate::b;
Improve this answer

3 Comments

Would it be possible to explain your answer in to the context? I am having trouble following your solution
I suppose this works only if you include b in the main.rs file with mod b.
This recipe fails with Rust 1.70.
4

In the latest Rust 1.63, we can use super keyword to refer siblings

// main.rs
mod a;
mod b;

fn main() {
}

// a.rs
pub fn a() {
    println!("A");
    super::b::b();
}

// b.rs
pub fn b() {
    println!("B");
}
Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.