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I have 4 files:

main.rs

mod bar;

fn main() {
    let v = vec![1, 2, 3];
    println!("Hello, world!");
}

lib.rs

pub mod foo;
pub mod bar;

foo.rs

pub fn say_foo() {

}

bar.rs

use crate::foo;

fn bar() {
    foo::say_foo();
}

When I run cargo run I get an error saying:

error[E0432]: unresolved import `crate::foo`
 --> src/bar.rs:1:5
  |
1 | use crate::foo;
  |     ^^^^^^^^^^ no `foo` in the root

Could someone explain to me how to fix this? A bit more broadly: how does module lookup work when there's a main.rs and a lib.rs?

Edit: Adding mod foo to main.rs fixes the issue. But I don't understand this -- I was under the impression the lib.rs was the place that "exposed" all of my modules? Why do I have to declare the module in main.rs as well?

My Cargo.toml:

[package]
name = "hello-world"
version = "0.1.0"
authors = ["me@mgail.com>"]
edition = "2018"

# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html

[dependencies]
nz_21
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  • What do you want to do? If bar needs foo to work, how do you want to use bar in the binary without foo? – Boiethios Sep 02 '19 at 12:31
  • What's in your `Cargo.toml`? Do you have [a library and a binary](https://stackoverflow.com/questions/26946646/rust-package-with-both-a-library-and-a-binary) in one project or not? Seems that you're mixing things together a bit. – zrzka Sep 02 '19 at 12:47

4 Answers4

312

Let's start from the beginning. Look at the Package Layout chapter in The Cargo Book. As you can see, your package can contain lot of stuff:

  • a binary (something you can run) or multiple binaries,
  • a single library (shared code),
  • example(s),
  • benchmark(s),
  • integration tests.

Package layout

Not all of the possibilities are listed here, just the binary / library combinations.

A binary

This is an example of a package with single binary. Entry point is the main function in the src/main.rs.

Cargo.toml:

[package]
name = "hallo"
version = "0.1.0"
edition = "2018"

src/main.rs:

fn main() {
    println!("Hallo, Rust here!")
}
$ cargo run
Hallo, Rust here!

A library

This is an example of a package with a library. Libraries don't have entry points, you can't run them. They're used for functionality sharing.

Cargo.toml:

[package]
name = "hallo"
version = "0.1.0"
edition = "2018"

src/lib.rs:

pub fn foo() {
    println!("Hallo, Rust library here!")
}
$ cargo run
error: a bin target must be available for `cargo run`

Do you see anything in the Cargo.toml file about a binary or a library? No. The reason is that I've followed the Package Layout and the cargo knows where to look for things.

A binary and a library

This is an example of a package with a binary and a library.

Cargo.toml:

[package]
name = "hallo"
version = "0.1.0"
edition = "2018"

src/lib.rs:

pub const GREETING: &'static str = "Hallo, Rust library here!";

src/main.rs:

use hallo::GREETING;

fn main() {
    println!("{}", GREETING);
}

Same question, do you see anything in the Cargo.toml file about a binary or a library? No.

This package contains two things:

  • a binary (root src/main.rs, entry point src/main.rs::main),
  • a library (root src/lib.rs, shared code).

A library can be referenced from the binary via use hallo::... where the hallo is this package name (Cargo.toml -> [package] -> name).

Your problem

Cargo.toml:

[package]
name = "hallo"
version = "0.1.0"
edition = "2018"

Same package layout

A library part

src/lib.rs:

pub mod bar;
pub mod foo;

src/foo.rs:

pub fn say_foo() {
    println!("Foo");
}

src/bar.rs:

use crate::foo;

pub fn bar() {
    foo::say_foo();
}

crate refers to src/lib.rs, because we're in the context of our library here.

Treat it as a standalone unit and refer to it via use hallo::...; from the outside world.

A binary part

src/main.rs:

use hallo::bar::bar;

fn main() {
    bar();
}

Here we're just using our library.

Without a library

Same code, but lib.rs was renamed to utils.rs and (foo|bar).rs files were moved to the src/utils/ folder.

src/utils.rs:

pub mod bar;
pub mod foo;

src/utils/foo.rs:

pub fn say_foo() {
    println!("Foo");
}

src/utils/bar.rs:

use super::foo;
// or use crate::utils::foo;

pub fn bar() {
    foo::say_foo();
}

We can use crate here as well, but because we're in the context of our binary, the path differs.

src/main.rs:

use utils::bar::bar;

mod utils;

fn main() {
    bar();
}

Here we just declared another module (utils) and we're using it.

Summary

Cargo.toml content:

[package]
name = "hallo"
version = "0.1.0"
edition = "2018"

If there's a src/main.rs file, you're basically saying this:

[package]
name = "hallo"
version = "0.1.0"
edition = "2018"

[[bin]]
name = "hallo"
src = "src/main.rs"

If there's a src/lib.rs file, you're basically saying this:

[package]
name = "hallo"
version = "0.1.0"
edition = "2018"

[lib]
name = "hallo"
path = "src/lib.rs"

If there're both of them, you're basically saying this:

[package]
name = "hallo"
version = "0.1.0"
edition = "2018"

[[bin]]
name = "hallo"
path = "src/main.rs"

[lib]
name = "hallo"
path = "src/lib.rs"

Documentation

zrzka
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  • "`crate` refers to `src/lib.rs`, because we're in the context of our library here." But hang on though, aren't `foo` and `bar` both in the same crate context, since they're both modules within `lib.rs`? Why would `bar` have to refer to `foo` as if it were external? – L.Y. Sim Sep 03 '19 at 13:45
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    You mean `use crate::foo;`? Why do you think it's an external reference, it's not. The `crate` keyword refers to [the current crate](https://doc.rust-lang.org/edition-guide/rust-2018/module-system/path-clarity.html#the-crate-keyword-refers-to-the-current-crate), which is the library itself. You can use `use super::foo;` if you wish. – zrzka Sep 03 '19 at 14:36
  • Hi, I believe there might have been some confusion because OP wasn't clear in the question. The `bar` module was declared in `lib.rs` and not in `main.rs`. Please see my answer to OP below for a more detailed response. – L.Y. Sim Sep 03 '19 at 14:41
  • No, the OP edited the question and tried to fix it by: _Adding mod foo to main.rs fixes the issue_. – zrzka Sep 03 '19 at 14:43
  • Actually never mind, I might have made a mistake myself – L.Y. Sim Sep 03 '19 at 14:43
  • Right it does appear that I was confused. Because OP was claiming that adding `mod bar` to `main.rs` solved the issue when in fact removing it is what will allow the code to compile since then `bar` will be properly declared as a module of `lib.rs` and `use crate::foo` can then properly refer to the `foo` module in `lib.rs`. – L.Y. Sim Sep 03 '19 at 14:53
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    Your post is confusing in the end. You are referring a src/lib.rs in your Cargo.toml while it was already renamed to utils.rs... – decades Oct 23 '20 at 07:30
  • @zrzka, thanks for this very details explanation, I create an example following the `Same package layout` part, still got the same error, https://github.com/videni/rust-package-layout-demo – Vidy Videni Apr 14 '21 at 03:12
  • Thanks, great answer. I had a co-located main.rs + lib.rs and was trying to use `crate::my_pub_fn` notation, swapping to `[crate_name]::my_pub_fn` was what I needed :) – Eats Indigo Feb 28 '23 at 00:41
59

In short the official Rust book has this to say:

If a package contains src/main.rs and src/lib.rs, it has two crates: a library and a binary, both with the same name as the package.

Furthermore the Rust reference says this:

crate resolves the path relative to the current crate

So there are actually two crates in your project, and to which crate the crate qualifier resolves to depends on where you call it.

Now in your code example, if you want things to compile you have to remove mod bar; from src/main.rs. Otherwise you'll be declaring that bar is a module within two crates.

After you remove that, then because in src/lib.rs you had:

pub mod foo;
pub mod bar;

bar would now be a module within src/lib.rs's crate, so the crate qualifier in bar.rs would then refer to src/lib.rs's hello-world crate, which is what you want.


One more thing, if you wanted to access items that are exposed in src/lib.rs from src/main.rs, you have to do as @zrzka said, which is to name the name of the crate that both src/lib.rs and src/main.rs share. For example, in your project which is named hello-world:

use hello_world::foo;
fn main() {
    foo::say_foo();
}

is how you import the foo module declared in src/lib.rs into src/main.rs.

However it does appear that the importing behavior doesn't work the other way. I.e. if you declare some public module in src/main.rs, you can't import it into the src/lib.rs crate even when you specify the name of the crate. I couldn't find documentation describing this behavior but by testing it in Rust 1.37.0, it does appear to be the case.

L.Y. Sim
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The lib.rs and main.rs files are two independent entry points for your package.

When you use cargo run (or build the binary and run it explicitly), the entry point to be used is main.rs, and the crate keyword refer to the binary crate. It doesn't even have to know that there is something in lib.rs: the binary will treat the library as it would any other external crate, and it must be imported, through extern crate hello_world or, for example, use hello_world::foo.

When you import the library, however, the entry point is lib.rs, and the crate is the library crate. In this case, yes, all that you've added to lib.rs is exposed to the whole crate.

The usual worksflow in this case is to make the binary something like a thin wrapper around the library - in some extreme cases the main.rs would only contain something like

fn main() {
    library::main();
}

and the whole logic (and all the project structure) goes into the library crate. One of the reasons is exactly what you've run into: the possible confusion whether this concrete module is imported in each crate in the package.

Kevin Reid
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Cerberus
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    This answer is very helpful. In other words, you can imagine that 1. `lib.rs` is really called `mod.rs`, 2. the folder `src` is really called `name_of_my_crate` as given by the `name` field of `Cargo.toml`'s `[package]` table, and 3. `main.rs` actually resides outside your crate (so you have to reference your crate by its name, not by `crate::`). – BallpointBen Nov 03 '21 at 04:07
0

One more thing: mod defines a new module in your crate, whether a binary crate or library crate; while use only brings the module into the current scope.

In your example, use crate::foo in bar.rs tries to bring a module named foo under crate root into scope. But because there is no mod foo in main.rs, the foo module is not part of the binary crate.

Silentroar
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