Going through some exercises on basic Rust syntax and ownership. Links: Exercises: https://101-rs.tweede.golf/A1-language-basics/mod.html, Slides: http://artificialworlds.net/presentations/rust-101...

cross-posted from: https://diode.zone/videos/watch/9766d1f1-6018-48ec-ad67-e971758f8a3a

Going through some exercises on basic Rust syntax and ownership.

Links:

Rust 101 is a series of videos explaining how to write programs in Rust. The course materials for this series are developed by tweede golf. You can find more information at https://github.com/tweedegolf/101-rs and you can sponsor the work at https://github.com/sponsors/tweedegolf . They are released under the Creative Commons Attribution Share Alike 4.0 International license.

This series of videos is copyright 2023 Andy Balaam and the tweede golf contributors and is released under the Creative Commons Attribution Share Alike 4.0 International license.

These videos are roughly on track with the Reading Club apparently, so this video belongs here this week, I think.

  • deur@feddit.nlEnglish
    3·
    9 months ago

    f32 is a type, you need to actually provide a value to the shorthand array initialization syntax, the array is filled with that value. There is no such thing as “uninitialized” or implicit zeroing here.

    let foo: [f32; 5] = [0.0; 5];

    You seem to have indicated dismay over the default f64 type.

    To type a integer or float literal, suffix it with its built-in type. For example:

    0.0f32

    1024usize

    Does this look ugly? Yes. Good news! It’s hardly relevant in an actual project because Rust will easily default to whichever float type you are using throughout the project as your floating point numbers with resolved types will resolve the types of the implicitly typed floating point numbers.

    You can also use the same syntax with the vec![] macro :)

    let bar = vec![0.0f32; 5];

    • Jayjader@jlai.luMEnglish
      3·
      9 months ago

      Shouldn’t we be able to do something like

      let bar : [f32; 5] = [Default::default(); 5];

      ?

      I’m on mobile and too lazy to try whipping up a rust playground example to test this out myself.

      • deur@feddit.nlEnglish
        3·
        9 months ago

        I’ll do you one better!

        let foo: [u8; 10] = Default::default();

        (pretty sure the above only works reliably after they finally were able to derive traits on n-sized arrays but that’s been in for a while now)

        And as you suggested:

        let foo: [u8; 10] = [Default::default(); 10];
    • maegul (he/they)MEnglish
      2·
      9 months ago

      Thanks!

      I recognise I was getting nitpicky there, just figured it might help someone else avoid confusion around it.

      And yea, I figured I’d make sure I knew how to set the numeric type of my choice rather than just rely on inference.

      Otherwise, I think I hate the postfix typing, I just can’t bring myself to use it lol.

      But thanks for the help!!