Should I call ResizeObserver.unobserve for removed elements?

Question

When some observed element gets removed from DOM, should I call .unobserve for that element to prevent memory leaks, or will it be "unobserved automatically"?

const ro = new ResizeObserver((entries) => { console.log(entries); });
const el = document.getElementById('foo');
ro.observe(el);
// ... some time later
el.remove();
ro.unobserve(el); // <-- is this needed, or does it happen automatically behind the scenes?

Why I'm asking: I'm implementing a React component that observes many children and properly cleaning up the observer for unmounted components would involve non-trivial code, which I want to avoid if it is actually unneeded.

Answer 1

According to the issue in the csswg-drafts repository, today Chrome automatically unobserves the element if you remove it from DOM and delete any references from JS.

let el = document.querySelector("#id");
let ro = new ResizeObserver();
ro.observe(el);
setTimeout(_ => {
  el.remove();
  el = null;
}

Current Chrome code implicitly unobserves element if there are no JS references to it. This is not captured by the specification.

But the issue is still open...