implement a list using Rust

Posted 2022-10-01 cutepig

Rust果然比較複雜，在經歷了n次compile fail，終于寫成了一個 list

難點： 對Rc<>的用法不熟悉。對borrow checker不夠熟悉

有些寫法可能還不是最短的

 

use std::rc::Rc;

fn main() 
    println!("Hello, world!");

    let li = ListInternal 
        data: 1,
        next: None,
    ;
    let li = ListInternal 
        data: 2,
        next: Some(Rc::new(li)),
    ;

    let mut l = List  next: Some(li) ;

    for i in 1..10
        l.push_back(i);
    
    
    l.print();


struct ListInternal 
    data: i32,
    next: Option<Rc<ListInternal>>,


struct List 
    next: Option<ListInternal>,


impl ListInternal 
    fn print(&self) 
        println!("", self.data);

        if let Some(ref v) = self.next 
            v.print();
        
    

    fn print_loop(&self) 
        let mut t = self;
        loop 
            println!("", t.data);

            if let Some(ref v) = t.next 
                t = v;
             else 
                break;
            
        
    

    fn last(&mut self) -> &mut Self 
        let mut t = self;
        loop 
            if let Some(ref mut v) = t.next 
                t = Rc::get_mut(v).unwrap();
             else 
                break;
            
        
        t
    

impl List 
    fn new() -> List 
        List  next: None 
    

    fn print(&self) 
        if let Some(ref v) = self.next 
            v.print_loop();
        
    
    fn push_back(&mut self, data: i32) 
        let li = ListInternal 
            data: data,
            next: None,
        ;

        if let Some(ref mut v) = self.next 
            v.last().next = Some(Rc::new(li));
         else 
            self.next = Some(li);