I'm trying to write a procedural macro that implements a #[derive()] trait for a struct. In the generated implementation code I need to use AnyMap.
To avoid multi crate imports, and based on what I read in some other crate's code (namely Serde) I put my generated code into a const _IMPL_xxx_FOR_xxx : () = { /* generated code */ }; block but it fails to compile.
I was able to replicate my issue with the following code
const BLOCK_1: () = {
extern crate anymap;
use anymap::AnyMap;
};
const BLOCK_2: () = {
extern crate anymap;
use anymap::AnyMap;
};
fn main() {
println!("foo");
}
The compile error I'm getting is the following:
error[E0432]: unresolved import `anymap::AnyMap`
--> src/main.rs:3:9
|
3 | use anymap::AnyMap;
| ^^^^^^^^^^^^^^ Maybe a missing `extern crate anymap;`?
error[E0432]: unresolved import `anymap::AnyMap`
--> src/main.rs:9:9
|
9 | use anymap::AnyMap;
| ^^^^^^^^^^^^^^ Maybe a missing `extern crate anymap;`?
Is it an issue specific to AnyMap? Would you know of any way to fix this (including maybe a different approach to generating procedural macro code if the pattern I'm using is not recommended?
I can replicate this on the playground using simply
const A: () = {
extern crate core;
use core::option::Option;
};
fn main() {}
However, it appears only the use statement is broken, and I can still use items from core, but I have to name them explicitly each time:
const A: () = {
extern crate core;
do_stuff!(core::option::Option)
};
The reason is that use statements typically assume a path relative to the root, and there is no way to explicitly name the block you are in (self refers to the current module unfortunately).
Here's a better workaround - as I said before, Rust uses self to refer to the current module, so you can just put a module in your code block and then have the use statements reference self.
In your case it would be:
const BLOCK_1: () = {
mod inner {
extern crate anymap;
use self::anymap::AnyMap;
}
};
On the specific issue of making this work for procedural macro, a suggested solution was to reexport the needed crate as part of the crate containing the derive macro (or the one containing the class I'm trying to export) using something like pub extern crate anymap and then use <my_crate>::anymap::AnyMap;
Related
I have two modules in separate files within the same crate, where the crate has macro_rules enabled. I want to use the macros defined in one module in another module.
// macros.rs
#[macro_export] // or not? is ineffectual for this, afaik
macro_rules! my_macro(...)
// something.rs
use macros;
// use macros::my_macro; <-- unresolved import (for obvious reasons)
my_macro!() // <-- how?
I currently hit the compiler error "macro undefined: 'my_macro'"... which makes sense; the macro system runs before the module system. How do I work around that?
Macros within the same crate
New method (since Rust 1.32, 2019-01-17)
foo::bar!(); // works
mod foo {
macro_rules! bar {
() => ()
}
pub(crate) use bar; // <-- the trick
}
foo::bar!(); // works
With the pub use, the macro can be used and imported like any other item. And unlike the older method, this does not rely on source code order, so you can use the macro before (source code order) it has been defined.
Old method
bar!(); // Does not work! Relies on source code order!
#[macro_use]
mod foo {
macro_rules! bar {
() => ()
}
}
bar!(); // works
If you want to use the macro in the same crate, the module your macro is defined in needs the attribute #[macro_use]. Note that macros can only be used after they have been defined!
Macros across crates
Crate util
#[macro_export]
macro_rules! foo {
() => ()
}
Crate user
use util::foo;
foo!();
Note that with this method, macros always live at the top-level of a crate! So even if foo would be inside a mod bar {}, the user crate would still have to write use util::foo; and not use util::bar::foo;. By using pub use, you can export a macro from a module of your crate (in addition to it being exported at the root).
Before Rust 2018, you had to import macro from other crates by adding the attribute #[macro_use] to the extern crate util; statement. That would import all macros from util. This syntax should not be necessary anymore.
Alternative approach as of 1.32.0 (2018 edition)
Note that while the instructions from #lukas-kalbertodt are still up to date and work well, the idea of having to remember special namespacing rules for macros can be annoying for some people.
EDIT: it turns out their answer has been updated to include my suggestion, with no credit mention whatsoever 😕
On the 2018 edition and onwards, since the version 1.32.0 of Rust, there is another approach which works as well, and which has the benefit, imho, of making it easier to teach (e.g., it renders #[macro_use] obsolete). The key idea is the following:
A re-exported macro behaves as any other item (function, type, constant, etc.): it is namespaced within the module where the re-export occurs.
It can then be referred to with a fully qualified path.
It can also be locally used / brought into scope so as to refer to it in an unqualified fashion.
Example
macro_rules! macro_name { ... }
pub(crate) use macro_name; // Now classic paths Just Work™
And that's it. Quite simple, huh?
Feel free to keep reading, but only if you are not scared of information overload ;) I'll try to detail why, how and when exactly does this work.
More detailed explanation
In order to re-export (pub(...) use ...) a macro, we need to refer to it! That's where the rules from the original answer are useful: a macro can always be named within the very module where the macro definition occurs, but only after that definition.
macro_rules! my_macro { ... }
my_macro!(...); // OK
// Not OK
my_macro!(...); /* Error, no `my_macro` in scope! */
macro_rules! my_macro { ... }
Based on that, we can re-export a macro after the definition; the re-exported name, then, in and of itself, is location agnostic, as all the other global items in Rust 🙂
In the same fashion that we can do:
struct Foo {}
fn main() {
let _: Foo;
}
We can also do:
fn main() {
let _: A;
}
struct Foo {}
use Foo as A;
The same applies to other items, such as functions, but also to macros!
fn main() {
a!();
}
macro_rules! foo { ... } // foo is only nameable *from now on*
use foo as a; // but `a` is now visible all around the module scope!
And it turns out that we can write use foo as foo;, or the common use foo; shorthand, and it still works.
The only question remaining is: pub(crate) or pub?
For #[macro_export]-ed macros, you can use whatever privacy you want; usually pub.
For the other macro_rules! macros, you cannot go above pub(crate).
Detailed examples
For a non-#[macro_export]ed macro
mod foo {
use super::example::my_macro;
my_macro!(...); // OK
}
mod example {
macro_rules! my_macro { ... }
pub(crate) use my_macro;
}
example::my_macro!(...); // OK
For a #[macro_export]-ed macro
Applying #[macro_export] on a macro definition makes it visible after the very module where it is defined (so as to be consistent with the behavior of non-#[macro_export]ed macros), but it also puts the macro at the root of the crate (where the macro is defined), in an absolute path fashion.
This means that a pub use macro_name; right after the macro definition, or a pub use crate::macro_name; in any module of that crate will work.
Note: in order for the re-export not to collide with the "exported at the root of the crate" mechanic, it cannot be done at the root of the crate itself.
pub mod example {
#[macro_export] // macro nameable at `crate::my_macro`
macro_rules! my_macro { ... }
pub use my_macro; // macro nameable at `crate::example::my_macro`
}
pub mod foo {
pub use crate::my_macro; // macro nameable at `crate::foo::my_macro`
}
When using the pub / pub(crate) use macro_name;, be aware that given how namespaces work in Rust, you may also be re-exporting constants / functions or types / modules. This also causes problems with globally available macros such as #[test], #[allow(...)], #[warn(...)], etc.
In order to solve these issues, remember you can rename an item when re-exporting it:
macro_rules! __test__ { ... }
pub(crate) use __test__ as test; // OK
macro_rules! __warn__ { ... }
pub(crate) use __warn__ as warn; // OK
Also, some false positive lints may fire:
from the trigger-happy clippy tool, when this trick is done in any fashion;
from rustc itself, when this is done on a macro_rules! definition that happens inside a function's body: https://github.com/rust-lang/rust/issues/78894
This answer is outdated as of Rust 1.1.0-stable.
You need to add #![macro_escape] at the top of macros.rs and include it using mod macros; as mentioned in the Macros Guide.
$ cat macros.rs
#![macro_escape]
#[macro_export]
macro_rules! my_macro {
() => { println!("hi"); }
}
$ cat something.rs
#![feature(macro_rules)]
mod macros;
fn main() {
my_macro!();
}
$ rustc something.rs
$ ./something
hi
For future reference,
$ rustc -v
rustc 0.13.0-dev (2790505c1 2014-11-03 14:17:26 +0000)
Adding #![macro_use] to the top of your file containing macros will cause all macros to be pulled into main.rs.
For example, let's assume this file is called node.rs:
#![macro_use]
macro_rules! test {
() => { println!("Nuts"); }
}
macro_rules! best {
() => { println!("Run"); }
}
pub fn fun_times() {
println!("Is it really?");
}
Your main.rs would look sometime like the following:
mod node; //We're using node.rs
mod toad; //Also using toad.rs
fn main() {
test!();
best!();
toad::a_thing();
}
Finally let's say you have a file called toad.rs that also requires these macros:
use node; //Notice this is 'use' not 'mod'
pub fn a_thing() {
test!();
node::fun_times();
}
Notice that once files are pulled into main.rs with mod, the rest of your files have access to them through the use keyword.
I have came across the same problem in Rust 1.44.1, and this solution works for later versions (known working for Rust 1.7).
Say you have a new project as:
src/
main.rs
memory.rs
chunk.rs
In main.rs, you need to annotate that you are importing macros from the source, otherwise, it will not do for you.
#[macro_use]
mod memory;
mod chunk;
fn main() {
println!("Hello, world!");
}
So in memory.rs you can define the macros, and you don't need annotations:
macro_rules! grow_capacity {
( $x:expr ) => {
{
if $x < 8 { 8 } else { $x * 2 }
}
};
}
Finally you can use it in chunk.rs, and you don't need to include the macro here, because it's done in main.rs:
grow_capacity!(8);
The upvoted answer caused confusion for me, with this doc by example, it would be helpful too.
Note: This solution does work, but do note as #ineiti highlighted in the comments, the order u declare the mods in the main.rs/lib.rs matters, all mods declared after the macros mod declaration try to invoke the macro will fail.
I'd like to write a server that resizes a huge image. Since loading it on every request would take a lot of time, I decided to pre-load it. Unfortunately I got the following error:
Compiling hello_world v0.0.0 (/tmp/Rocket/examples/hello_world)
error[E0015]: calls in statics are limited to constant functions, tuple structs and tuple variants
--> examples/hello_world/src/main.rs:9:35
|
9 | static img: image::DynamicImage = image::open("/home/d33tah/tmp/combined.png").unwrap();
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
error[E0015]: calls in statics are limited to constant functions, tuple structs and tuple variants
--> examples/hello_world/src/main.rs:9:35
|
9 | static img: image::DynamicImage = image::open("/home/d33tah/tmp/combined.png").unwrap();
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
error: aborting due to 2 previous errors
For more information about this error, try `rustc --explain E0015`.
error: Could not compile `hello_world`.
To learn more, run the command again with --verbose.
Here's the code:
#![feature(proc_macro_hygiene, decl_macro)]
#[macro_use] extern crate rocket;
#[cfg(test)] mod tests;
extern crate image;
static img: image::DynamicImage = image::open("/home/d33tah/tmp/combined.png").unwrap();
#[get("/")]
fn hello() -> Vec<u8> {
"".into()
}
fn main() {
rocket::ignite().mount("/", routes![hello]).launch();
}
To compile it, you'll need rocket installed based on its latest Github checkout (currently: 831d8dfbe30dd69f0367a75361e027127b2100e1) and image crate.
Is it possible to create such a global variable? If not, do I have any other options?
EDIT: this was flagged as a duplicate of this question, but as Boiethios showed, this specific case is better solved by Rocket's API.
If you want to share something in Rocket, you will have to use a managed state:
Say to Rocket you want to manage the resource:
let image = image::open("/home/d33tah/tmp/combined.png").unwrap();
rocket::ignite().manage(image);
Retrieve it in the routes where you need it:
#[get("/foo")]
fn foo(image: State<DynamicImage>) {
// Can use `image`.
}
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've seen a couple tutorials to create a Python module using the cpython crate but still have errors when building:
extern crate cpython;
use cpython::{PyObject, PyResult, Python, PyTuple, PyDict, ToPyObject, PythonObject};
fn add_two(py: Python, args: &PyTuple, _: Option<&PyDict>) -> PyResult<PyObject> {
match args.as_slice() {
[ref a_obj, ref b_obj] => {
let a = a_obj.extract::<i32>(py).unwrap();
let b = b_obj.extract::<i32>(py).unwrap();
let mut acc:i32 = 0;
for _ in 0..1000 {
acc += a + b;
}
Ok(acc.to_py_object(py).into_object())
},
_ => Ok(py.None())
}
}
py_module_initializer!(example, |py, module| {
try!(module.add(py, "add_two", py_fn!(add_two)));
Ok(())
});
I get:
error: macro undefined: 'py_module_initializer!'
Where do I get it? I am using Rust 1.12.
UPD
Need to add #[macro_use] (as in answers)
For other errors - see examples here
You probably need to declare cpython as follows:
#[macro_use] extern crate cpython;
To be able to use cpython's macros. You can consult the example in its docs.
You must add the #[macro_use] attribute on the extern crate declaration to ask the compiler to bring the macros exported by the crate into your crate's namespace.
#[macro_use]
extern crate cpython;
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.