How can I (de)serialize object that has trait field to .ron format with an erased-serde crate?

the erased-serde example for json and cbor.

https://github.com/dtolnay/erased-serde

how can i change object with trait fields to ron format str?

[dependencies]
ron = "0.8"
serde = "1.0"
erased-serde = "0.3"

use serde::Serialize;

trait Temp {
    fn temp(&self);
}

struct TempNum {
    num: i32,
}

impl Temp for TempNum {
    fn temp(&self) {}
}

#[derive(Serialize)]
struct IncludeTrait {
    temps: Vec<Box<dyn Temp>>, // this is problem
}

fn main() {
    let object = IncludeTrait { temps: vec![Box::new(TempNum { num: 123 })], };
    let str = ron::to_string(&object).unwrap();
    let object: IncludeTrait = ron::from_str(&str).unwrap();
}

To use erased-serde, you need to declare erased_serde::Serialize as a supertrait of your trait and implement Serialize for dyn Temp via that (you can do that with erased_serde::serialize_trait_object!()):

use serde::Serialize;

trait Temp: erased_serde::Serialize {
    fn temp(&self);
}

erased_serde::serialize_trait_object!(Temp);

#[derive(Serialize)]
struct TempNum {
    num: i32,
}

impl Temp for TempNum {
    fn temp(&self) {}
}

#[derive(Serialize)]
struct IncludeTrait {
    temps: Vec<Box<dyn Temp>>,
}

However, even that helps only with serialization, and from your code it looks like you also want to support deserialization. Deserialization of trait object is complicated - we can't know what object to construct. But we have the typetag crate for that. Just add #[typetag::serde] on top of the trait and the impls:

use serde::{Deserialize, Serialize};

#[typetag::serde]
trait Temp {
    fn temp(&self);
}

#[derive(Serialize, Deserialize)]
struct TempNum {
    num: i32,
}

#[typetag::serde]
impl Temp for TempNum {
    fn temp(&self) {}
}

#[derive(Serialize, Deserialize)]
struct IncludeTrait {
    temps: Vec<Box<dyn Temp>>,
}