For the following JavaScript code, how can I write it in ReasonML?
class HelloWorld extends HTMLElement {
constructor() {
super();
// Attach a shadow root to the element.
let shadowRoot = this.attachShadow({mode: 'open'});
shadowRoot.innerHTML = `<p>hello world</p>`;
}
}
I could not find any documentation on writing classes in ReasonML? I cannot use plain objects/types as I need to extend from HTMLElement class which doesn't work with ES style classes.
I have looked into this existing question - How to extend JS class in ReasonML however, it is a different thing. To write web component, we need to extend HTMLElement and must call it with new keyword. ES5 style extension mechanism doesn't work.
You can't. Not directly at least, since BuckleScript (which Reason uses to compile to JavaScript) targets ES5 and therefore has no knowledge of ES6 classes.
Fortunately, ES6-classes require no special runtime support, but are implemented as just syntax sugar, which is why you can transpile ES6 to ES5 as shown in the question you link to. All you really have to do then, is to convert this transpiled output into ReasonML:
var __extends = (this && this.__extends) || function (d, b) {
for (var p in b) if (b.hasOwnProperty(p)) d[p] = b[p];
function __() { this.constructor = d; }
d.prototype = b === null ? Object.create(b) : (__.prototype = b.prototype, new __());
};
var BaseElement = (function (_super) {
__extends(BaseElement, _super);
function BaseElement() {
_super.call(this);
}
return BaseElement;
}(HTMLElement));
And depending on what specific class-features you actually need, you can probably simplify it a bit.
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.
What should i do if i want to export some ScalaJS methods as CommonJS module? I have the following but it doesn't seem to work:
#ScalaJSDefined
#JSExportTopLevel("default")
object SourceFetch extends js.Object {
def activate(state: js.Dynamic): Unit = {
global.console.log("activate")
}
def deactivate(): Unit = {
global.console.log("deactivate")
}
}
And yes, scalaJSModuleKind := ModuleKind.CommonJSModule is in the build.sbt.
What i want as output is a commonjs module that looks like this;
export default {
activate(state) {
console.log("activate");
}.
deactivate() {
console.log("deactivate");
}
};
What i ended up doing is to use the deprecated sbt key "scalaJSOutputWrapper" and append 'module.exports = exports["default"];' at the end of output JS file.
I did try "scalaJSUseMainModuleInitializer" but I am only able to get a hold of "module.exports" not "exports" and the value of "module.exports" is undefined.
Your above snippet does indeed correspond to the piece of ECMAScript 2015 code that you wrote. However, that does not export the methods as direct members of the module, but as members of the default object. And no, the default export is not the same as the module itself (although many people think so).
To export the functions as direct members of the module, you should write:
object SourceFetch {
#JSExportTopLevel("activate")
def activate(state: js.Dynamic): Unit = {
global.console.log("activate")
}
#JSExportTopLevel("deactivate")
def deactivate(): Unit = {
global.console.log("deactivate")
}
}
I'm using IntelliJ IDEA's File Watcher to automatically compile the TypeScript files, but for some reason it's not liking classes defined within blocks / function closures:
Is there a way around this without having to move everything to the top-level / global scope?
Using the following code in TypeScript results in practically the same JavaScript that you appear to be aiming for...
namespace MY_NAMESPACE {
export class AssetService {
}
}
Resulting code:
var MY_NAMESPACE;
(function (MY_NAMESPACE) {
var AssetService = (function () {
function AssetService() {
}
return AssetService;
}());
MY_NAMESPACE.AssetService = AssetService;
})(MY_NAMESPACE || (MY_NAMESPACE = {}));
If you want to really reduce the scope, switch to external modules (AKA "modules" these days).
If you don't export the class from the module/file, it won't be visible globally, i.e. there's no reason to enclose class definitions in function scopes.
More about modules in TS: https://www.typescriptlang.org/docs/handbook/modules.html
Is there a way to do an alias or "use" (like PHP) for a TypeScript class/module.
Example:
If I have:
module Foo {
class Bar {}
}
Normally I have to write Foo.Bar to use it outside of the module. Is there a way I can alias that to something else, like "FooBar".
This would be really useful if you have several submodules (which my current project does), like:
module A.B.C.D {
export class E {}
}
is normally A.B.C.D.E which is silly.
According to page 82 of the old Typescript language spec, it stated that the following is possible.
So you should be able to alias "use" a module without having to reference the entire hierarchy.
module A.B.C
{
import XYZ = X.Y.Z;
export function ping(x: number) {
if (x > 0) XYZ.pong(x – 1);
}
}
module X.Y.Z
{
import ABC = A.B.C;
export function pong(x: number) {
if (x > 0) ABC.ping(x – 1);
}
}
The new Typescript language spec covers similar material in the section on Import Alias Declarations