如何序列化或反序列化Serde中的Arc <T>？

Question

我有一个包含自己类型的子的结构.这些孩子被包裹在Arcs中,我在调用serde_json::to_string它时遇到问题.我的结构是:

#[derive(Serialize, Deserialize)]
pub struct Category {
    pub id: i32,
    pub name: String,
    pub parent_id: i32,
    pub children: Vec<Arc<Category>>,
}

这会产生错误the trait 'serde::Serialize' is not implemented for 'std::sync::Arc<db::queries::categories::Category>'我尝试了一些不同的方法来使序列化工作,例如:

#[serde(serialize_with = "arc_category_vec")]
pub children: Vec<Arc<Category>>

fn arc_category_vec<S>(value: &Vec<Arc<Category>>, serializer: S) -> Result<S::Ok, S::Error>
where
    S: Serializer,
{
    let mut seq = serializer.serialize_seq(Some(value.len()))?;
    for e in value {
        seq.serialize_element(e.as_ref())?;
    }
    seq.end()
}

这没有帮助,因为我得到相同的错误.我也尝试过:

impl Serialize for Arc<Category> {
    fn serialize<S>(&self, serializer: S) -> Result<S::Ok, S::Error>
    where
        S: Serializer,
    {
        let mut state = serializer.serialize_struct("Category", 4)?;

        state.serialize_field("id", &self.id)?;
        state.serialize_field("name", &self.name)?;
        state.serialize_field("parent_id", &self.parent_id)?;
        state.serialize_field("children", &self.children)?;
        state.end();
    }
}

但这给出了错误 impl doesn't use types inside crate

我可能没有反序列化,因为序列化在这一点上更重要.

Answer 1

SERDE提供的实现Serialize和Deserialize对Arc<T>和Rc<T>,但只有当rc功能被启用.

Serde的参考网站上有评论解释原因:

选择投入Rc<T>和投注Arc<T>.序列化和反序列化这些类型不会保留标识,并可能导致相同数据的多个副本.在启用此功能之前,请确保这是您想要的.

要启用此rc功能,您需要自己提出要求Cargo.toml:

[dependencies]
serde = { version = "1.0", features = ["rc"] }