Stabilise inline_const by nbdd0121 · Pull Request #104087 · rust-lang/rust
github.com
Stabilisation Report Summary This PR will stabilise inline_const feature in expression position. inline_const_pat is still unstable and will not be stabilised. The feature will allow code like this...
  • kevincox@lemmy.ml
    6·
    8 months ago

    This is a nice small feature. I’m curious about the commit description:

    foo(const { 1 + 1 })

    which is roughly desugared into

    struct Foo;
impl Foo {
    const FOO: i32 = 1 + 1;
}
foo(Foo::FOO)

    I would have expected it to desugar to something like:

    foo({
  const TMP: i32 = 1 + 1;
  TMP
})

    But I can’t seem an explanation why the struct with impl is used. I wonder if it has something to do with propagating generics.

    • 0v0
      10·
      8 months ago

      It’s because it has to work in pattern contexts as well, which are not expressions.

        • 0v0
          10·
          8 months ago
          fn foo(x: i32) {
    match x {
        const { 3.pow(3) } => println!("three cubed"),
        _ => {}
    }
}

          But it looks like inline_const_pat is still unstable, only inline_const in expression position is now stabilized.

    • Giooschi@lemmy.worldEnglish
      7·
      8 months ago

      They tested the same strings on that implementation

      The code they were looking at was used for writing the table, but they were testing the one that read it (which is instead correct).

      though judging by the recent comments someone’s found something.

      Yeah that’s me :)The translation using an associated const also works when the const block uses generic parameters. For example:

      fn require_zst<T>() {
    const { assert!(std::mem::size_of::<T>() == 0) }
}

      This can be written as:

      fn require_zst<T>() {
    struct Foo<T>(PhantomData<T>);
    impl<T> Foo<T> {
        const FOO: () = assert!(std::mem::size_of::<T>() == 0);
    }
    Foo::<T>::FOO
}

      However it cannot be written as:

      fn require_zst<T>() {
    const FOO: () = assert!(std::mem::size_of::<T>() == 0);
    FOO
}

      Because const FOO: () is an item, thus it is only lexically scoped (i.e. visible) inside require_zst, but does not inherit its generics (thus it cannot use T).