I've met a conflict with Rust's ownership rules and a trait object downcast. This is a sample:
use std::any::Any;
trait Node{
fn gen(&self) -> Box<Node>;
}
struct TextNode;
impl Node for TextNode{
fn gen(&self) -> Box<Node>{
Box::new(TextNode)
}
}
fn main(){
let mut v: Vec<TextNode> = Vec::new();
let node = TextNode.gen();
let foo = &node as &Any;
match foo.downcast_ref::<TextNode>(){
Some(n) => {
v.push(*n);
},
None => ()
};
}
The TextNode::gen method has to return Box<Node> instead of Box<TextNode>, so I have to downcast it to Box<TextNode>.
Any::downcast_ref's return value is Option<&T>, so I can't take ownership of the downcast result and push it to v.
====edit=====
As I am not good at English, my question is vague.
I am implementing (copying may be more precise) the template parser in Go standard library.
What I really need is a vector, Vec<Box<Node>> or Vec<Box<Any>>, which can contain TextNode, NumberNode, ActionNode, any type of node that implements the trait Node can be pushed into it.
Every node type needs to implement the copy method, return Box<Any>, and then downcasting to the concrete type is OK. But to copy Vec<Box<Any>>, as you don't know the concrete type of every element, you have to check one by one, that is really inefficient.
If the copy method returns Box<Node>, then copying Vec<Box<Node>> is simple. But it seems that there is no way to get the concrete type from trait object.