I have a little project and I want to encapsulate a struct's fields and use implemented methods.
├── src
├── main.rs
├── predator
└── prey
├── cycle.rs
└── mod.rs
cycle.rs
struct Prey {
name: String,
}
impl Prey {
pub fn new(n: String) -> Prey {
Prey { name: n }
}
pub fn get_name(&self) -> &str {
self.name.as_str()
}
}
I'd like to leave Prey as private.
main.rs
use prey::cycle::Prey;
mod prey;
fn main() {
let pr = Prey::new("Hamster".to_string());
println!("Hello, world! {}", pr.get_name());
}
I get an error:
error: struct `Prey` is private
I know that if I put pub before struct Prey {}, I'll get the expected result. I will be grateful for an explanation, how, why not and/or best practices.
Visibility works at the module level. If you want module X to have access to an item in module Y, module Y must make it public.
Modules are items too. If you don't make a module public, then it's internal to your crate. Therefore, you shouldn't worry about making items in that module public; only your crate will have access to it.
The crate root (usually the file named lib.rs or main.rs) is the root module of your crate. It defines the public interface of the crate, i.e. public items in the crate root are accessible from other crates.
In your example, you write mod prey;. That defines the prey module as private, so items in the prey module are not accessible from other crates (unless you reexport them with pub use). That means you should make prey::cycle::Prey public.
Related
This is my directory structure,
.
├── Cargo.lock
├── Cargo.toml
├── src
│ ├── add.rs
│ └── main.rs
└── tests
└── add_test.rs
add.rs
pub fn add(a: u8, b: u8) -> u8 {
return a + b;
}
main.rs
pub mod add;
fn main() {}
add_test.rs
#[cfg(test)]
mod tests {
#[test]
fn add_test() {
// how do I use add.rs module?
}
}
In add_test function how do I test add.rs's add function?
Notice that you are using mod add;. This is not a public module meaning it is unavailable outside of your crate.
To have it available outside the crate (and therefore available in the test) you can either make the module public, or reexport the function itself.
Additionally, to do this, you will need a lib.rs file in src where the exports are put:
// Make module public
pub mod add;
// Make the function available at the root of the crate
pub use add::add;
To then use your function in tests, you call it the same way you would call a function for a different crate. Lets assume your crate was named your-crate:
// If your module is public
your-crate::add::add(2,3);
// If you reexport the function
your-crate::add(2,3);
See more details in the chapter on test organization in the Rust book.
This question already has answers here:
How can I include a module from another file from the same project?
(6 answers)
Closed 1 year ago.
I created a Cargo project. My project tree looks like this:
├── src
├── main.rs
├── animal.rs
└── human.rs
animal.rs
use crate::human::Job;
pub struct Animal {
id: u8,
job: Job,
}
impl Animal {
pub fn new(i: u8, j: Job) -> Animal {
Animal {id:i, job:j}
}
}
human.rs
pub struct Human {
id: u8,
job: Job,
}
pub enum Job {
Builder,
Magician,
}
impl Human {
pub fn new(i: u8, j: Job) -> Human {
Human {id: i, job: j}
}
}
When I compile it, it complains
$ cargo run
error[E0433]: failed to resolve: could not find `human` in the crate root
--> src/animal.rs:4:17
|
4 | job: crate::human::Job,
| ^^^^^ could not find `human` in the crate root
It gets compiled if I add "mod human;" in main.rs. What is the right way to include one module from another file in the same project?
The default package structure cargo expects is described here: https://doc.rust-lang.org/cargo/guide/project-layout.html
Therefore, you would expect to have the following files:
└── src
├── main.rs
├── lib.rs
├── animal.rs
└── human.rs
And lib.rs should contain
mod animal;
mod human;
The main.rs is an executable that would use the crate as if the library was a dependency:
use <crate_name>::human::Job;
fn main() {
// do whatever with `Job`
}
crate_name here would be the name of the crate in Cargo.toml.
Although technically nothing prevents you from declaring the modules inside main.rs like so:
mod animal;
mod human;
fn main() {
// use `human::Job`
}
I am trying to share a struct between two files, but I am getting an error.
I have the following folder structure:
src/
Models/
Login.rs
Routes/
LoginRoute.rs
Services/
LoginService.rs
main.rs
In Login.rs I have:
#[derive(Serialize, Deserialize, Debug)]
pub struct UserLoginResponse {
id: i32,
username: String,
token: String
}
In LoginRoute.rs I have:
#[path = "../Models/Login.rs"]
pub mod Login;
#[path = "../Services/LoginService.rs"]
pub mod LoginService;
#[post("/login", format = "application/json", data = "<user>")]
pub async fn login(user: String) -> Json<Login::UserLoginResponse> {
if let Ok(sk) = LoginService::callAuthenticate(user).await {
return sk
......
In LoginService.rs I have:
#[path = "../Models/Login.rs"]
pub mod Login;
pub async fn callAuthenticate(user: String)-> Result<Json<Login::UserLoginResponse>, Error> {
...
let userLoginResponse :Login::UserLoginResponse = Login::UserLoginResponse::new(1, "admin".to_string(), api_reponse.return_result);
Ok(Json(userLoginResponse))
}
I am getting error in LoginRoute.rs on the return sk line:
expected struct 'LoginRoute::Login::UserLoginResponse', found struct 'LoginService::Login:UserLoginResponse'
Please do not use the #[path = ...] attribute for your typical organization; it should only be used in obscure cases. Each time you do mod something, you are declaring a new module. Even if two modules point to the same file due to #[path = ...], they will be distinct.
So you have multiple UserLoginResponse structs declared:
one at crate::LoginRoute::Login::UserLoginResponse
one at crate::LoginService::Login:UserLoginResponse
and maybe another if you've also declared Login in main.rs.
Since they're in distinct modules, the Rust compiler sees them as different types, which is not what you want.
Just use the idiomatic way of declaring modules. If you want to keep your existing folder structure without intermediate mod.rs files, you can declare them all in main.rs like so:
mod Models {
pub mod Login;
}
mod Routes {
pub mod LoginRoute;
}
mod Services {
pub mod LoginService;
}
And then access them elsewhere via crate::Models::Login and whatnot.
See:
How do I import from a sibling module?
You've probably already run into warnings from the compiler trying to encourage a specific style: "module [...] should have a snake case name". Idiomatic file structure would typically look like this:
src/
models/
login.rs
mod.rs
routes/
login_route.rs
mod.rs
services/
login_service.rs
mod.rs
main.rs
Where main.rs would have:
mod models;
mod routes;
mod services;
And src/models/mod.rs (for example) would have:
pub mod login;
I am trying to write a crate called bar, the structure looks like this
src/
├── bar.rs
└── lib.rs
My src/lib.rs looks like this
#![crate_type = "lib"]
#![crate_name = "bar"]
#![feature(ip_addr)]
#[allow(dead_code)]
pub mod bar;
My bar.rs has
pub struct baz {
// stuff
}
impl baz {
// stuff
}
Now when I try to use this crate in another crate like:
extern crate bar;
use bar::baz;
fn main() {
let cidr = baz::new("Hi");
println!("{}", cidr.say());
}
This fails with
error: unresolved import `bar::baz`. There is no `baz` in `bar`
Do I need to declare the module somewhere else?
The important part you are missing is that crates define their own module. That is, your crate bar implicitly defines a module called bar, but you also have created a module called bar inside that. Your struct resides within this nested module.
If you change your main to use bar::bar::baz; you can progress past this. You will have to decide if that's the structure you want though. Most idiomatic Rust projects would not have the extra mod and would flatten it out:
src/lib.rs
pub struct Baz {
// stuff
}
impl Baz {
// stuff
}
Unfortunately, your example code cannot compile, as you have invalid struct definitions, and you call methods that don't exist (new), so I can't tell you what else it will take to compile.
Also, structs should be PascalCase.
If you have a directory structure like this:
src/main.rs
src/module1/blah.rs
src/module1/blah2.rs
src/utils/logging.rs
How do you use functions from other files?
From the Rust tutorial, it sounds like I should be able to do this:
main.rs
mod utils { pub mod logging; }
mod module1 { pub mod blah; }
fn main() {
utils::logging::trace("Logging works");
module1::blah::doit();
}
logging.rs
pub fn trace(msg: &str) {
println!(": {}\n", msg);
}
blah.rs
mod blah2;
pub fn doit() {
blah2::doit();
}
blah2.rs
mod utils { pub mod logging; }
pub fn doit() {
utils::logging::trace("Blah2 invoked");
}
However, this produces an error:
error[E0583]: file not found for module `logging`
--> src/main.rs:1:21
|
1 | mod utils { pub mod logging; }
| ^^^^^^^
|
= help: name the file either logging.rs or logging/mod.rs inside the directory "src/utils"
It appears that importing down the path, i.e. from main to module1/blah.rs works, and importing peers, i.e. blah2 from blah works, but importing from the parent scope doesn't.
If I use the magical #[path] directive, I can make this work:
blah2.rs
#[path="../utils/logging.rs"]
mod logging;
pub fn doit() {
logging::trace("Blah2 invoked");
}
Do I really have to manually use relative file paths to import something from a parent scope level? Isn't there some better way of doing this in Rust?
In Python, you use from .blah import x for the local scope, but if you want to access an absolute path you can use from project.namespace.blah import x.
I'm going to answer this question too, for anyone else who finds this and is (like me) totally confused by the difficult-to-comprehend answers.
It boils down to two things I feel are poorly explained in the tutorial:
The mod blah; syntax imports a file for the compiler. You must use this on all the files you want to compile.
As well as shared libraries, any local module that is defined can be imported into the current scope using use blah::blah;.
A typical example would be:
src/main.rs
src/one/one.rs
src/two/two.rs
In this case, you can have code in one.rs from two.rs by using use:
use two::two; // <-- Imports two::two into the local scope as 'two::'
pub fn bar() {
println!("one");
two::foo();
}
However, main.rs will have to be something like:
use one::one::bar; // <-- Use one::one::bar
mod one { pub mod one; } // <-- Awkwardly import one.rs as a file to compile.
// Notice how we have to awkwardly import two/two.rs even though we don't
// actually use it in this file; if we don't, then the compiler will never
// load it, and one/one.rs will be unable to resolve two::two.
mod two { pub mod two; }
fn main() {
bar();
}
Notice that you can use the blah/mod.rs file to somewhat alleviate the awkwardness, by placing a file like one/mod.rs, because mod x; attempts x.rs and x/mod.rs as loads.
// one/mod.rs
pub mod one.rs
You can reduce the awkward file imports at the top of main.rs to:
use one::one::bar;
mod one; // <-- Loads one/mod.rs, which loads one/one.rs.
mod two; // <-- This is still awkward since we don't two, but unavoidable.
fn main() {
bar();
}
There's an example project doing this on Github.
It's worth noting that modules are independent of the files the code blocks are contained in; although it would appear the only way to load a file blah.rs is to create a module called blah, you can use the #[path] to get around this, if you need to for some reason. Unfortunately, it doesn't appear to support wildcards, aggregating functions from multiple files into a top-level module is rather tedious.
I'm assuming you want to declare utils and utils::logging at the top level, and just wish to call functions from them inside module1::blah::blah2. The declaration of a module is done with mod, which inserts it into the AST and defines its canonical foo::bar::baz-style path, and normal interactions with a module (away from the declaration) are done with use.
// main.rs
mod utils {
pub mod logging { // could be placed in utils/logging.rs
pub fn trace(msg: &str) {
println!(": {}\n", msg);
}
}
}
mod module1 {
pub mod blah { // in module1/blah.rs
mod blah2 { // in module1/blah2.rs
// *** this line is the key, to bring utils into scope ***
use crate::utils;
pub fn doit() {
utils::logging::trace("Blah2 invoked");
}
}
pub fn doit() {
blah2::doit();
}
}
}
fn main() {
utils::logging::trace("Logging works");
module1::blah::doit();
}
The only change I made was the use crate::utils; line in blah2 (in Rust 2015 you could also use use utils or use ::utils). Also see the second half of this answer for more details on how use works. The relevant section of The Rust Programming Language is a reasonable reference too, in particular these two subsections:
Separating Modules into Different Files
Bringing Paths into Scope with the use Keyword
Also, notice that I write it all inline, placing the contents of foo/bar.rs in mod foo { mod bar { <contents> } } directly, changing this to mod foo { mod bar; } with the relevant file available should be identical.
(By the way, println(": {}\n", msg) prints two new lines; println! includes one already (the ln is "line"), either print!(": {}\n", msg) or println!(": {}", msg) print only one.)
It's not idiomatic to get the exact structure you want, you have to make one change to the location of blah2.rs:
src
├── main.rs
├── module1
│ ├── blah
│ │ └── blah2.rs
│ └── blah.rs
└── utils
└── logging.rs
main.rs
mod utils {
pub mod logging;
}
mod module1 {
pub mod blah;
}
fn main() {
utils::logging::trace("Logging works");
module1::blah::doit();
}
utils/logging.rs
pub fn trace(msg: &str) {
println!(": {}\n", msg);
}
module1/blah.rs
mod blah2;
pub fn doit() {
blah2::doit();
}
module1/blah/blah2.rs (the only file that requires any changes)
// this is the only change
// Rust 2015
// use utils;
// Rust 2018
use crate::utils;
pub fn doit() {
utils::logging::trace("Blah2 invoked");
}
I realize this is a very old post and probably wasn't using 2018. However, this can still be really tricky and I wanted to help those out that were looking.
Because Pictures are worth a thousand words I made this simple for code splitting.
Then as you probably guessed they all have an empty pub fn some_function().
We can further expand on this via the changes to main
The additional changes to nested_mod
Let's now go back and answer the question:
We added blah1 and blah2 to the mod_1
We added a utils with another mod logging inside it that calls some fn's.
Our mod_1/mod.rs now contains:
pub mod blah.rs
pub mod blah2.rs
We created a utils/mod.rs used in main containing:
pub mod logging
Then a directory called logging/with another mod.rs where we can put fns in logging to import.
Source also here https://github.com/DavidWhit/Rust_Modules
Also Check Chapters 7 for libs example and 14.3 that further expands splitting with workspaces in the Rust Book. Good Luck!
Answers here were unclear for me so I will put my two cents for future Rustaceans.
All you need to do is to declare all files via mod in src/main.rs (and fn main of course).
// src/main.rs
mod module1 {
pub mod blah;
pub mod blah2;
}
mod utils {
pub mod logging;
}
fn main () {
module1::blah::doit();
}
Make everything public with pub you need to use externally (copy-pasted original src/utils/logging.rs). And then simply use declared modules via crate::.
// src/module1/blah.rs
use crate::utils::logging;
// or `use crate::utils` and `utils::logging::("log")`, however you like
pub fn doit() {
logging::trace("Logging works");
}
p.s. I shuffled functions a bit for a cleaner answer.
If you create a file called mod.rs, rustc will look at it when importing a module. I would suggest that you create the file src/utils/mod.rs, and make its contents look something like this:
pub mod logging;
Then, in main.rs, add a statement like this:
use utils::logging;
and call it with
logging::trace(...);
or you could do
use utils::logging::trace;
...
trace(...);
Basically, declare your module in the mod.rs file, and use it in your source files.