Let's say I have the following main.rs and lib.rs files. Why after doing the wildcard import self refers to testing::nested and not the main binary module anymore?
Could somebody point me to documentation where they explain this? I haven't been able to find it.
// main.rs
use testing::nested::*;
fn main() {
self::hello();
}
// lib.rs (testing module)
pub mod nested {
pub hello() {
println!("hello from nested::hello");
}
}
Just like #Ivan and #eggyal mentioned, is not that self refers to nested, is that the use declaration created a binding for hello in the main module.
Related
I must admit, I am having a hard time making sense of the Rust compiler error messages.
I have this module:
src/adapters.js
use actix_web::{Responder, HttpResponse};
pub mod Basic {
pub fn api_index() -> &'static str {
"API"
}
pub fn admin_index() -> impl Responder {
HttpResponse::Ok().body("hello world!")
}
}
The Rust compiler keeps telling me when I use
use crate::actix_web::{Responder, HttpResponse};
that:
E0432: unresolved import `crate::actix_web` maybe a missing crate `actix_web`?
Rest assured, crate::actix_web is not missing, because I can run simple requests from main.rs. The problem starts when I want to use the actix_web modules within my own modules.
If I have
use actix_web::{Responder, HttpResponse};
Rust keeps teeling me:
error[E0433]: failed to resolve: use of undeclared type or module `HttpResponse`
HttpResponse::Ok().body("hello world!")
^^^^^^^^^^^^ not found in this scope
help: consider importing one of these items
use actix_web::HttpResponse;
use crate::adapters::HttpResponse;
E0432: unresolved import `crate::actix_web` maybe a missing crate `actix_web`?
I my main.rs, as the first line, I've also declared:
extern crate actix_web;
For the sake of making sure, I've also added "extern crate actix_web;" to the adapters.rs module right at the start. That didn't change anything either.
I am running out of options, I don't know what else I could possibly do.
Basically You have created one another module Basic inside adapters module.So Basic is submodule of adapters module. So two ways you can access the external crate in this case.
Import external crate modules inside your Basic Module.
pub mod Basic {
use actix_web::{Responder, HttpResponse};
pub fn api_index() -> &'static str {
"API"
}
pub fn admin_index() -> impl Responder {
HttpResponse::Ok().body("hello world!")
}
}
Or import an external module in outer module module and access those module using use super keyword in inner module.
use actix_web::{Responder, HttpResponse};
pub mod Basic {
pub fn api_index() -> &'static str {
"API"
}
pub fn admin_index() -> impl super::Responder {
super::HttpResponse::Ok().body("hello world!")
}
}
I have the following code:
pub mod a {
#[test]
pub fn test() {
println!("{:?}", std::fs::remove_file("Somefilehere"));
}
}
I get errors when I compile this:
error[E0433]: failed to resolve. Use of undeclared type or module `std`
--> src/main.rs:4:24
|
4 | println!("{}", std::fs::remove_file("Somefilehere"));
| ^^^ Use of undeclared type or module `std`
However, removing the inner module and compiling the code it contains by itself works fine:
#[test]
pub fn test() {
println!("{:?}", std::fs::remove_file("Somefilehere"));
}
What am I missing here? I get the same errors if the module is in a separate file:
main.rs
pub mod a;
a.rs
#[test]
pub fn test() {
println!("{:?}", std::fs::remove_file("Somefilehere"));
}
By default, the compiler inserts extern crate std; at the beginning of the crate root (the crate root is the file that you pass to rustc). This statement has the effect of adding the name std to the crate's root namespace and associating it with a module that contains the public contents of the std crate.
However, in child modules, std is not automatically added in the module's namespace. This is why the compiler cannot resolve std (or anything that starts with std::) in a module.
There are many ways to fix this. First, you can add use std; in a module to make the name std refer, within that module, to the root std. Note that in use statements, the path is treated as absolute (or "relative to the crate's root namespace"), whereas everywhere else, paths are treated as relative to the current namespace (be it a module, a function, etc.).
pub mod a {
use std;
#[test]
pub fn test() {
println!("{:?}", std::fs::remove_file("Somefilehere"));
}
}
You can also use a use statement to import more specific items. For example, you can write use std::fs::remove_file;. This lets you avoid having to type the whole path to remove_file and just use the name remove_file directly within that module:
pub mod a {
use std::fs::remove_file;
#[test]
pub fn test() {
println!("{:?}", remove_file("Somefilehere")));
}
}
Finally, you can avoid using use altogether by prefixing the path with :: to ask the compiler to resolve the path from the crate's root namespace (i.e. turning the path into an absolute path).
pub mod a {
#[test]
pub fn test() {
println!("{:?}", ::std::fs::remove_file("Somefilehere"));
}
}
The recommended practice is to import types (structs, enums, etc.) directly (e.g. use std::rc::Rc;, then use the path Rc), but to use functions through an import of their parent module (e.g. use std::io::fs;, then use the path fs::remove_file).
pub mod a {
use std::fs;
#[test]
pub fn test() {
println!("{:?}", fs::remove_file("Somefilehere"));
}
}
Side note: You can also write self:: at the beginning of a path to make it relative to the current module. This is more often used in use statements, since other paths are already relative (though they are relative to the current namespace, whereas self:: is always relative to the containing module).
Nowadays, std is directly accessible from everywhere, so the code you showed is compiling as you would expect.
Furthermore, extern crate is no longer needed in Rust edition 2018. Adding a dependency to Cargo.toml makes the crate name directly available as a top-level identifier.
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.
I have the following code:
pub mod a {
#[test]
pub fn test() {
println!("{:?}", std::fs::remove_file("Somefilehere"));
}
}
I get errors when I compile this:
error[E0433]: failed to resolve. Use of undeclared type or module `std`
--> src/main.rs:4:24
|
4 | println!("{}", std::fs::remove_file("Somefilehere"));
| ^^^ Use of undeclared type or module `std`
However, removing the inner module and compiling the code it contains by itself works fine:
#[test]
pub fn test() {
println!("{:?}", std::fs::remove_file("Somefilehere"));
}
What am I missing here? I get the same errors if the module is in a separate file:
main.rs
pub mod a;
a.rs
#[test]
pub fn test() {
println!("{:?}", std::fs::remove_file("Somefilehere"));
}
By default, the compiler inserts extern crate std; at the beginning of the crate root (the crate root is the file that you pass to rustc). This statement has the effect of adding the name std to the crate's root namespace and associating it with a module that contains the public contents of the std crate.
However, in child modules, std is not automatically added in the module's namespace. This is why the compiler cannot resolve std (or anything that starts with std::) in a module.
There are many ways to fix this. First, you can add use std; in a module to make the name std refer, within that module, to the root std. Note that in use statements, the path is treated as absolute (or "relative to the crate's root namespace"), whereas everywhere else, paths are treated as relative to the current namespace (be it a module, a function, etc.).
pub mod a {
use std;
#[test]
pub fn test() {
println!("{:?}", std::fs::remove_file("Somefilehere"));
}
}
You can also use a use statement to import more specific items. For example, you can write use std::fs::remove_file;. This lets you avoid having to type the whole path to remove_file and just use the name remove_file directly within that module:
pub mod a {
use std::fs::remove_file;
#[test]
pub fn test() {
println!("{:?}", remove_file("Somefilehere")));
}
}
Finally, you can avoid using use altogether by prefixing the path with :: to ask the compiler to resolve the path from the crate's root namespace (i.e. turning the path into an absolute path).
pub mod a {
#[test]
pub fn test() {
println!("{:?}", ::std::fs::remove_file("Somefilehere"));
}
}
The recommended practice is to import types (structs, enums, etc.) directly (e.g. use std::rc::Rc;, then use the path Rc), but to use functions through an import of their parent module (e.g. use std::io::fs;, then use the path fs::remove_file).
pub mod a {
use std::fs;
#[test]
pub fn test() {
println!("{:?}", fs::remove_file("Somefilehere"));
}
}
Side note: You can also write self:: at the beginning of a path to make it relative to the current module. This is more often used in use statements, since other paths are already relative (though they are relative to the current namespace, whereas self:: is always relative to the containing module).
Nowadays, std is directly accessible from everywhere, so the code you showed is compiling as you would expect.
Furthermore, extern crate is no longer needed in Rust edition 2018. Adding a dependency to Cargo.toml makes the crate name directly available as a top-level identifier.