• kevincox@lemmy.ml
    link
    fedilink
    arrow-up
    6
    ·
    7 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@sopuli.xyz
      link
      fedilink
      arrow-up
      10
      ·
      7 months ago

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

        • 0v0@sopuli.xyz
          link
          fedilink
          arrow-up
          10
          ·
          7 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.world
      link
      fedilink
      English
      arrow-up
      7
      ·
      7 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).

  • sugar_in_your_tea@sh.itjust.works
    link
    fedilink
    arrow-up
    6
    arrow-down
    1
    ·
    7 months ago

    Huh, this is awesome! From the link, this now works:

    let v: Vec<i32> = const { Vec::new() };
    

    I’m going to have to play with this to see how far it goes.

    • asdfasdfasdf@lemmy.world
      link
      fedilink
      arrow-up
      2
      ·
      edit-2
      7 months ago

      What I’m curious about - this function was already const, so for stuff like this, I’d think there’s basically a 100% chance the compiler would optimize this too, just implicitly.

      AFAIK this new feature is just for times when it isn’t an optimization, but more your own domain invariants. E.g. assertions.

      But I could be wrong. I wonder if this can be used for actual optimizations in some places that the compiler couldn’t figure out by itself.

      • sugar_in_your_tea@sh.itjust.works
        link
        fedilink
        arrow-up
        2
        ·
        edit-2
        7 months ago

        Ah, apparently for now you’re not allowed to allocate. But vec::new_in(allocator) looks interesting. This works in nightly today:

        #![feature(allocator_api)]
        
        use std::alloc::Global;
        
        fn main() {
            const MY_VEC: Vec<i32> = const {
                Vec::new_in(Global)
            };
            println!("{:?}", MY_VEC);
        }
        

        Maybe at some point I can append to it at compile time too. I’d love to be able to put a const {} and have allocations that resolve down to a 'static, and this seems to be a step toward that.

        I guess I’m just excited that Vec::new() is the example they picked, since the next obvious question is, “can I push?”

  • onlinepersona@programming.dev
    link
    fedilink
    English
    arrow-up
    2
    arrow-down
    2
    ·
    edit-2
    7 months ago

    I’m not getting it. What’s the point? It seems very much like a cpp-ism where you can put const in so many places.

    const int n2 = 0;    // const object
    int const n3 = 0;    // const object (same as n2)
    // https://learn.microsoft.com/en-us/cpp/cpp/const-and-volatile-pointers?view=msvc-170
    const char *cpch;  // const variable cannot point to another pointer
    char * const pchc; // value of pointer is constant
    
    int f() const; // members cannot be modified in this, only read
    std::string const f(); // returns a constant
    

    Then there are constant expressions.

    Can anybody look at that and tell me it’s readable with a straight face? I hope they don’t start adding all this stuff to rust.

    Anti Commercial-AI license

    • Miaou@jlai.lu
      link
      fedilink
      arrow-up
      4
      ·
      7 months ago

      It can be used for producing const values in arbitrary context. Can basically be swapped for c++'s constexpr.

      C++'s const does not exist in rust (values are const by default).

    • Alex@programming.dev
      link
      fedilink
      arrow-up
      3
      ·
      6 months ago

      Nope. This little neat feature mainly is just necessary part of bigger one - const-generics with const bounds.