Is there a way to return a function?

[jjangga0214]

Hi!

I'd like to declare a function whose return type is another function.

But the compiler complains.

Is there a proper valid way?

One workaround is this.

#[napi]
pub struct Foo;

#[napi]
impl Foo {
  #[napi]
  pub fn foo(&self) {
    println!("foo");
  }
}

#[napi]
pub fn foo() -> Foo {
  Foo {}
}

import { foo,bar } from './index.js'

const res = foo()
res.foo()

But this requires consumer to call res.foo(), not res().

To solve this issue, this solution may be used.

#![feature(unboxed_closures)]
#![feature(fn_traits)]

#[napi]
pub struct Foo;

#[napi]
impl Fn for Foo {
  #[napi]
  pub fn call(&self, _args: ()) {
    println!("foo");
  }
}

#[napi]
pub fn foo() -> Foo {
  Foo {}
}

import { foo } from './index.js'

const res = foo()
res()

But this requires #![feature] flag, which means depending on unstable version of Rust.

And I feel the first and second solutions are all boilerplate.

I just want to simply write this.

#[napi]
pub fn foo() {
  Box::new(|| println!("foo"))
}

Is there a way?

Thanks :)

[Brooooooklyn]

See this for example: https://github.com/napi-rs/napi-rs/blob/main/examples/napi/src/callback.rs#L51

[jjangga0214]

Thank you!
However, is there a way to return lambda?