Rust lifetimes for struct references - rust

Question

I've just started with Rust but can't quite grasp lifetimes so I could resolve following issue by myself:
This test project is about simulating a bit to allow tracing it through various bitwise operations, e.g. let newbit = oldbit1 ^ oldbit2 and looking at newbit I can tell afterwards it came out of an XOR operation with oldbit1 and oldbit2 as operands.
#[derive(Copy,Clone)]
pub enum TraceOperation {
AND,
OR,
XOR,
NOT,
}
#[derive(Copy,Clone)]
pub struct TraceBit<'a> {
source_a: Option<&'a TraceBit<'a>>,
source_b: Option<&'a TraceBit<'a>>,
source_op: Option<TraceOperation>,
value: bool,
}
This compiles, but I don't fully understand why the lifetime parameters are needed that way. I assume that the compiler cannot expect that the members source_a and source_b live as long as the struct itself as this may not hold true, so explicit lifetimes are required.
is this assumption correct?
Further I don't fully understand why I have to re-specify the lifetime parameter for the reference type, i.e. why I have to write source_a: Option<&'a TraceBit<'a>> as opposed to source_a: Option<&'a TraceBit>.
What is the second lifetime used for? How do I read that line out loud? I have:
"source_a is a variable of type Option that may have Some reference (that is valid at least as long as the struct itself and as long as member source_b) to an instance of TraceBit"
My final issue is that I cannot make it to work using an overloaded operator:
use std::ops::BitXor;
impl<'a> BitXor for TraceBit<'a> {
type Output = Self;
fn bitxor(self, rhs: Self) -> Self {
let valA: usize = if self.value { 1 } else { 0 };
let valB: usize = if rhs.value { 1 } else { 0 };
let val = if valA ^ valB != 0 { true } else { false };
TraceBit { source_a: Some(&self), source_b: Some(&rhs), source_op: Some(TraceOperation::XOR), value: val }
}
}
This is basically pure guessing based on BitXor documentation. So what I try to do, in a very explicit manner, is to perform an xor operation on the two input variables and create a new TraceBit as output with the inputs stored in it as reference.
error[E0597]: `self` does not live long enough
--> libbittrace/src/lib.rs:37:30
|
37 | TraceBit { source_a: Some(&self), source_b: Some(&rhs), source_op: Some(TraceOperation::XOR), value: val }
| ^^^^ does not live long enough
38 | }
| - borrowed value only lives until here
|
note: borrowed value must be valid for the lifetime 'a as defined on the impl at 31:1...
--> libbittrace/src/lib.rs:31:1
|
31 | / impl<'a> BitXor for TraceBit<'a> {
32 | | type Output = Self;
33 | | fn bitxor(self, rhs: Self) -> Self {
34 | | let valA: usize = if self.value { 1 } else { 0 };
... |
40 | |
41 | | }
| |_^
error[E0597]: `rhs` does not live long enough
--> libbittrace/src/lib.rs:37:53
|
37 | TraceBit { source_a: Some(&self), source_b: Some(&rhs), source_op: Some(TraceOperation::XOR), value: val }
| ^^^ does not live long enough
38 | }
| - borrowed value only lives until here
|
note: borrowed value must be valid for the lifetime 'a as defined on the impl at 31:1...
--> libbittrace/src/lib.rs:31:1
|
31 | / impl<'a> BitXor for TraceBit<'a> {
32 | | type Output = Self;
33 | | fn bitxor(self, rhs: Self) -> Self {
34 | | let valA: usize = if self.value { 1 } else { 0 };
... |
40 | |
41 | | }
| |_^
error: aborting due to 2 previous errors
Seems like nothing lives longer than the xor operation itself, but how can I resolve this?
I've tried various workarounds/changes to the code but to no avail and in any way I rather like to understand the issue than guessing a correct solution....

Answer 1

Tree-like structures must use the Box pointer type (Option<Box<TraceBit>>). In general, in structs you should prefer owned types.
Rust references aren't mere pointers. They are borrows (compile-time read/write locks) of data that must exist as owned somewhere else.
So if you have an owned version of TraceBit:
pub struct TraceBit {
source_a: Option<Box<TraceBit>>,
}
then reference to it is of type: &'a TraceBit, but references to a type don't change how the type looks internally, so the type of source_a is still Box<TraceBit>. You can keep getting the &'a TraceBit references recursively step by step:
trace_bit = trace_bit.source_a.as_ref().unwrap();
but there's no construct in Rust where taking a reference to the root of a tree suddenly changes the whole tree into a tree of references, so the type you are creating can't exist, and that's why you can't get type annotations right.

Answer 2

Maybe instead of passing references around, you should use a contained and cloneable name type.
use std::rc::Rc;
#[derive(Debug)]
pub enum TraceOperation {
AND,
OR,
XOR,
NOT,
}
#[derive(Debug)]
pub enum BitName<T> {
Name(Rc<T>),
Combination(Rc<(TraceOperation, BitName<T>, BitName<T>)>),
}
impl<T> Clone for BitName<T> {
fn clone(&self) -> Self {
match self {
&BitName::Name(ref x) => BitName::Name(Rc::clone(x)),
&BitName::Combination(ref x) => BitName::Combination(Rc::clone(x)),
}
}
}
impl<T> From<T> for BitName<T> {
fn from(x:T) -> Self {
BitName::Name(Rc::new(x))
}
}
impl<T> BitName<T> {
pub fn combine(op : TraceOperation, a : &Self, b :&Self) -> Self {
BitName::Combination(Rc::new((op, (*a).clone(), (*b).clone())))
}
}
fn main() {
let x : BitName<String> = BitName::from(String::from("x"));
let y : BitName<String> = BitName::from(String::from("y"));
let xandy = BitName::combine(TraceOperation::AND, &x, &y);
println!("{:?}", xandy);
}