AstPtr can de-sync from mutable Syntax Tree after mutation

[Technohacker]

If I keep AstPtrs to two nodes in a tree (appearing one after the other in source code), and make changes to the first one, the second AstPtr becomes invalid and will panic when to_node is called

In retrospect I should've seen this coming considering AstPtr only keeps a source code location reference to the node in the tree, and the tree source physically gets changed when mutations occur

Pseudocode to demonstrate the issue:

println!("Create Tree");
let mut tree = make_module("root");

println!("Add 2 children");
tree.append_child(make_module("child_1"));
tree.append_child(make_module("child_2"));

println!("Create child pointers");
let child_1_ptr = AstPtr::new(&tree.ast_children().nth(0).unwrap());
let child_2_ptr = AstPtr::new(&tree.ast_children().nth(1).unwrap());

println!("Resolve both pointers");
let mut child_1 = child_1_ptr.to_node(tree.syntax());
let child_2 = child_2_ptr.to_node(tree.syntax()); // All fine here, no mutations done

println!("Mutate tree under child 1");
child_1.append_child(make_module("child_1_1")).unwrap();

println!("Try to resolve pointer 2");
let child_2 = child_2_ptr.to_node(tree.syntax()); // Panic!

I'm fairly certain this is by design, but is there any way around this? If not, I could make a PR to add this as a note to the docs

[Veykril]

I think it's an error to allow making AstPtrs to mutable trees, since they easily fail as you've shown. We should note that down in the docs (and ideally assert that the node is from an immutable tree).