I have been using Windows for 30 years and Linux for 25 years (debian since 99’). I really would not bash (pun intended) windows users so much, there is place for both of them.

It’s insane to find a windows user that doesn’t live in the terminal, it’s just not designed for it

Linux has a gui for everything

The first directory block is a hole. But type == DIRENT, so no error is reported. After that, we get a directory block without ‘.’ and ‘…’ but with a valid dentry. This may cause some code that relies on dot or dotdot (such as make_indexed_dir()) to crash

The problem isn’t that the block is a hole. It’s that the downstream function expects the directory block to contain . and .., and it gets given one without because of incorrect error handling.

You can encode the invariant of “has dot and dot dot” using a refinement type and smart constructor. The refined type would be a directory block with a guarantee it meets that invariant, and an instance of it could only be created through a function that validates the invariant. If the invariant is met, you get the refined type. If it isn’t, you only get an error.

This doesn’t work in C, but in languages with stricter type systems, refinement types are a huge advantage.

What kind of type signature would prove the first block of any directory in an ext4 filesystem image isn’t a hole?

I don’t know if the type system proves it’s not a hole, but the type system certainly seems to force consumers to contend with the possibility by surfacing the outcomes at the type system level. That’s what the Either is doing in the example’s return type, is it not?

fn get_or_create_inode(
    &self,
    ino: Ino
) -> Result<Either<ARef<Inode<T>>, inode::New<T>>>

It doesn’t help that the Rust community tends to bring extremely divisive politics with it in places and ways that just don’t need to happen, starting battles that aren’t even tangentially related to programming.

Arrogant hypocrites are a pet peeve of mine. If someone is going to act like progressive technology changes are beneath them and unnecessary, they should be able to put their money where their mouth is.

Let me know when you find one?

Indeed :)

Even if you wrote the code yourself you can come back to it a while later and have a wtf moment ¯\_(ツ)_/¯

Also, even if you think it’s a bug it might be a feature that other people use and “fixing” changing it might break systems.

Gotta be mm to make sense unfortunately, Linux-GUN 7.62.11

I’ll try :) Looks like I still have my code from when I was grinding through The Book, and there’s a couple spots that might be illuminating from a pedagogical standpoint. That being said, I’m sure my thought process, and “what was active code and what was commented out and when,” will probably be hard to follow.

My first confusion was in deref coercion auto dereferencing (edit: see? it’s still probably not 100% in my head :P), and my confusion pretty much matched this StackOverflow entry:

https://stackoverflow.com/questions/28519997/what-are-rusts-exact-auto-dereferencing-rules

It took me until Chapter 15 of The Book (on Boxes) to really get a feel for what was happening. My work and comments for Chapter 15:

use crate::List::{Cons, Nil};
use std::ops::Deref;

enum List {
    Cons(i32, Box<List>),
    Nil,
}

struct MyBox<T>(T);

impl<T> Deref for MyBox<T> {
    type Target = T;
    fn deref(&self) -> &Self::Target {
        &self.0
    }
}

impl<T> MyBox<T> {
    fn new(x: T) -> MyBox<T> {
        MyBox(x)
    }
}

#[derive(Debug)]
struct CustomSmartPointer {
    data: String,
}

impl Drop for CustomSmartPointer {
    fn drop(&mut self) {
        println!("Dropping CustomSmartPointer with data `{}`!", self.data);
    }
}

fn main() {
    let b = Box::new(5);
    println!("b = {}", b);

    let _list = Cons(1, Box::new(Cons(2, Box::new(Cons(3,Box::new(Nil))))));

    let x = 5;
    let y = MyBox::new(x);

    assert_eq!(5,x);
    assert_eq!(5, *y);

    let m = MyBox::new(String::from("Rust"));
    hello(&m);
    hello(m.deref());
    hello(m.deref().deref());
    hello(&(*m)[..]);
    hello(&(m.deref())[..]);
    hello(&(*(m.deref()))[..]);
    hello(&(*(m.deref())));
    hello((*(m.deref())).deref());

    // so many equivalent ways. I think I'm understanding what happens
    // at various stages though, and why deref coercion was added to
    // the language. Would cut down on arguing over which of these myriad
    // cases is "idomatic." Instead, let the compiler figure out if there's
    // a path to the desired end state (&str).

    // drop stuff below ...
    let _c = CustomSmartPointer {
        data: String::from("my stuff"),
    };
    let _d = CustomSmartPointer {
        data: String::from("other stuff"),
    };

    println!("CustomSmartPointers created.");
    drop(_c);
    println!("CustomSmartPointer dropped before the end of main.");

    // this should fail.
    //println!("{:?}", _c);
    // yep, it does.

}

fn hello(name: &str) {
    println!("Hello, {name}!");
}

Another thing that ended up biting me in the ass was Non-Lexical Lifetimes (NLLs). My code from Chapter 8 (on HashMaps):

use std::collections::HashMap;

fn print_type_of<T>(_: &T) {
    println!("{}", std::any::type_name::<T>())
}

fn main() {
    let mut scores = HashMap::new();
    scores.insert(String::from("Red"), 10);
    scores.insert(String::from("Blue"), 20);

    let score1 = scores.get(&String::from("Blue")).unwrap_or(&0);
    println!("score for blue is {score1}");
    print_type_of(&score1); //&i32
    let score2 = scores.get(&String::from("Blue")).copied().unwrap_or(0);
    println!("score for blue is {score2}");
    print_type_of(&score2); //i32

    // hmmm... I'm thinking score1 is a "borrow" of memory "owned" by the
    // hashmap. What if we modify the blue teams score now? My gut tells
    // me the compiler would complain, since `score1` is no longer what
    // we thought it was. But would touching the score of Red in the hash
    // map still be valid? Let's find out.

    // Yep! The below two lines barf!
    //scores.insert(String::from("Blue"),15);
    //println!("score for blue is {score1}");

    // But can we fiddle with red independently?
    // Nope. Not valid. So... the ownership must be on the HashMap as a whole,
    // not pieces of its memory. I wonder if there's a way to make ownership
    // more piecemeal than that.
    //scores.insert(String::from("Red"),25);
    //println!("score for blue is {score1}");

    // And what if we pass in references/borrows for the value?
    let mut refscores = HashMap::new();
    let mut red_score:u32 = 11;
    let mut blue_score:u32 = 21;
    let default:u32 = 0;
    refscores.insert(String::from("red"),&red_score);
    refscores.insert(String::from("blue"),&blue_score);

    let refscore1 = refscores.get(&String::from("red")).copied().unwrap_or(&default);
    println!("refscore1 is {refscore1}");

    // and then update the underlying value?
    // Yep. This barfs, as expected. Can't mutate red_score because it's
    // borrowed inside the HashMap.
    //red_score = 12;
    //println!("refscore1 is {refscore1}");

    // what if we have mutable refs/borrows though? is that allowed?
    let mut mutrefscores = HashMap::new();
    let mut yellow_score:u32 = 12;
    let mut green_score:u32 = 22;
    let mut default2:u32 = 0;
    mutrefscores.insert(String::from("yellow"),&mut yellow_score);
    mutrefscores.insert(String::from("green"),&mut green_score);
    //println!("{:?}", mutrefscores);

    let mutrefscore1 = mutrefscores.get(&String::from("yellow")).unwrap();//.unwrap_or(&&default2);
    //println!("{:?}",mutrefscore1);
    
    println!("mutrefscore1 is {mutrefscore1}");

    // so it's allowed. But do we have the same "can't mutate in two places"
    // rule? I think so. Let's find out.

    // yep. same failure as before. makes sense.
    //yellow_score = 13;
    //println!("mutrefscore1 is {mutrefscore1}");

    // updating entries...
    let mut update = HashMap::new();
    update.insert(String::from("blue"),10);
    //let redscore = update.entry(String::from("red")).or_insert(50);
    update.entry(String::from("red")).or_insert(50);
    //let bluescore = update.entry(String::from("blue")).or_insert(12);
    update.entry(String::from("blue")).or_insert(12);


    //println!("redscore is {redscore}");
    //println!("bluescore is {bluescore}");
    println!("{:?}",update);

    // hmmm.... so we can iterate one by one and do the redscore/bluescore
    // dance, but not in the same scope I guess.
    let mut updatesingle = HashMap::new();
    updatesingle.insert(String::from("blue"),10);
    for i in "blue red".split_whitespace() {
        let score = updatesingle.entry(String::from(i)).or_insert(99);
        println!("score is {score}");
    }

    // update based on contents
    let lolwut = "hello world wonderful world";
    let mut lolmap = HashMap::new();
    for word in lolwut.split_whitespace() {
        let entry = lolmap.entry(word).or_insert(0);
        *entry += 1;
    }

    println!("{:?}",lolmap);

    // it seems like you can only borrow the HashMap as a whole.
    // let's try updating entries outside the context of a forloop.

    let mut test = HashMap::new();
    test.insert(String::from("hello"),0);
    test.insert(String::from("world"),0);
    let hello = test.entry(String::from("hello")).or_insert(0);
    *hello += 1;
    let world = test.entry(String::from("world")).or_insert(0);
    *world += 1;

    println!("{:?}",test);

    // huh? Why does this work? I'm borrowing two sections of the hashmap like before in the update
    // section.
    
    // what if i print the actual hello or world...
    // nope. barfs still.
    //println!("hello is {hello}");

    // I *think* what is happening here has to do with lifetimes. E.g.,
    // when I introduce the println macro for hello variable, the lifetime
    // gets extended and "crosses over" the second borrow, violating the
    // borrow checker rules. But, if there is no println macro for the hello
    // variable, then the lifetime for each test.entry is just the line it
    // happens on.
    //
    // Yeah. Looks like it has to do with Non-Lexical Lifetimes (NLLs), a
    // feature since 2018. I've been thinking of lifetimes as lexical this
    // whole time. And before 2018, that was correct. Now though, the compiler
    // is "smarter."
    //
    // https://stackoverflow.com/questions/52909623/rust-multiple-mutable-borrowing
    //
    //   https://stackoverflow.com/questions/50251487/what-are-non-lexical-lifetimes
    //let 
}