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This is my singly linked list code have to implement pop from end and pop from front functions still, but I'm having difficulty in writing pop from end function

#![allow(warnings)] #[derive(Clone)] struct List { el: i32, next: Box<Option<List>> } impl List { fn new(val: i32) -> List { return List { el: val, next: Box::new(None) } } fn from(arr: &[i32]) -> List { let mut head = List::new(arr[0]); let mut current = &mut head; for &val in &arr[1..] { /* current.next = Box::new(Some(List::new(val))); current = (*current.next).as_mut().unwrap(); */ current.append(val); } return head } fn append(&mut self, val: i32) { let mut current = self; while let Some(ref mut next_node) = *current.next { current = next_node; } current.next = Box::new(Some(List::new(val))); } fn prepend(&mut self, val:i32) { // std::mem::replace(dest, src): // Moves src into the referenced dest, returning the previous dest value let old_self = std::mem::replace(self, List::new(val)); self.next = Box::new(Some(old_self)); } } fn main() { let mut list = List::new(42); list.append(32); list.append(2321); list.append(2839); list.prepend(69); list.pop(); // 2839 // difficulty in implementing this println!("{}", list); let mut list = List::from(&[1, 2, 3]); list.prepend(0); println!("{}", list); }

Should I use Rc and RefCell to implement a singly linked list? I'm a beginner, so please help. Thanks!

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  • Please edit your question to add the minimal reproducible example that uses pop and the actual error message. Commented Jan 13 at 7:18
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    1. Don't use linked lists. 2. If you really must use a linked list, don't write it yourself. 3. If you still want to write a linked list, read Learning Rust With Entirely Too Many Linked Lists. Commented Jan 13 at 7:34
  • You have removed too much code, this is not a minimal reproducible example. How can we help you without seeing enough of the whole picture? Commented Jan 13 at 11:59
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    The votes are no judgement of your person, it is a judgement of your question. In its current state, it is not helpful for others with a similar issue. Therefore, I'll keep my DV. Commented Jan 13 at 12:00
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    This site is not a forum, and by far not the only resource for beginners. In contrast to forums, it means to be "a library of detailed, high-quality answers to every question about programming." (From the introducing statement of the tour, which you took some months ago.) Commented Jan 13 at 13:45

1 Answer 1

Reset to default
1

Possible solution

 fn pop(&mut self) -> Option<List> { match &mut*self.next { // if self.next is None // that means that the list only have one element // and idk what's you desired behavior, but I just return None None => None, // we check the next element Some(n) => { if n.next.is_none() { // if the next element is the terminal element // we take the self.next // that will change the current.next to None self.next.take() }else{ // if the next element is not the terminal element n.pop() } } } } 
fn main() { let mut list = List::new(42); list.append(32); list.append(2321); list.append(2839); list.prepend(69); println!("{}", list); println!("{:?}", list.pop()); println!("{}", list); let mut list = List::from(&[1, 2, 3]); list.prepend(0); println!("{}", list); println!("{:?}",list.pop()); println!("{}", list); }
Output
LinkedList([69, 42, 32, 2321, 2839]) Some(List { el: 2839, next: None }) LinkedList([69, 42, 32, 2321]) LinkedList([0, 1, 2, 3]) Some(List { el: 3, next: None }) LinkedList([0, 1, 2])
Remark
ref deref box

there is this pattern &*self.next

let x: Box<Option<List>> = self.next; // illegal, you cannot move it out let x: Option<List> = * self.next; // illegal, you cannot move it out let x: &Option<List> = & * self.next; // legal, you can borrow the value
derive debug

you can use #[derive(Debug)] to make debug easier, and don't need to implement Display all the time.

#[derive(Clone, Debug)] struct List { el: i32, next: Box<Option<List>> }
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