在泛型结构上派生反序列化时无法解析 T: serde::Deserialize<'a>

Question

我正在尝试编写一个派生的结构，serde::Deserialize但它也有一个应该派生的字段serde::Deserialize：

extern crate serde;
#[macro_use]
extern crate serde_derive;

use serde::{Deserialize, Serialize};

#[derive(PartialEq, Serialize, Deserialize)]
pub struct Record<'a, T>
where
    T: 'a + Serialize + Deserialize<'a>,
{
    id: &'a str,
    created_at: &'a str,
    created_by: Option<&'a str>,
    last_updated_at: Option<&'a str>,
    object: &'a T,
}

impl<'a, T> Record<'a, T>
where
    T: 'a + Serialize + Deserialize<'a>,
{
    pub fn new(
        id: &'a str,
        created_at: &'a str,
        created_by: Option<&'a str>,
        last_updated_at: Option<&'a str>,
        object: &'a T,
    ) -> Self {
        Record {
            id,
            created_at,
            created_by,
            last_updated_at,
            object,
        }
    }
}

fn main() {}

我一直在更改代码一段时间，但我无法编译这个想法。我现在得到的错误是：

extern crate serde;
#[macro_use]
extern crate serde_derive;

use serde::{Deserialize, Serialize};

#[derive(PartialEq, Serialize, Deserialize)]
pub struct Record<'a, T>
where
    T: 'a + Serialize + Deserialize<'a>,
{
    id: &'a str,
    created_at: &'a str,
    created_by: Option<&'a str>,
    last_updated_at: Option<&'a str>,
    object: &'a T,
}

impl<'a, T> Record<'a, T>
where
    T: 'a + Serialize + Deserialize<'a>,
{
    pub fn new(
        id: &'a str,
        created_at: &'a str,
        created_by: Option<&'a str>,
        last_updated_at: Option<&'a str>,
        object: &'a T,
    ) -> Self {
        Record {
            id,
            created_at,
            created_by,
            last_updated_at,
            object,
        }
    }
}

fn main() {}

Answer 1

通常，您不应在 structs 上编写 Serde 特征边界。

rustc --explain E0283 解释你的问题：

当编译器没有足够的信息来明确选择实现时会发生此错误

我发现#[serde(bound()]用于声明边界使示例编译：

#[derive(PartialEq, Serialize, Deserialize)]
pub struct Record<'a, T: 'a> {
    id: &'a str,
    created_at: &'a str,
    created_by: Option<&'a str>,
    last_updated_at: Option<&'a str>,
    #[serde(bound(deserialize = "&'a T: Deserialize<'de>"))]
    object: &'a T,
}

作为另一种解决方案，因为T是通用的并且可能是参考，请考虑更改Record定义，以便 Serde 不需要更明确的指示：

#[derive(PartialEq, Serialize, Deserialize)]
pub struct Record<'a, T: 'a> {
    id: &'a str,
    created_at: &'a str,
    created_by: Option<&'a str>,
    last_updated_at: Option<&'a str>,
    object: T,
}

impl<'a, T: 'a> Record<'a, T> {
    pub fn new(
        id: &'a str,
        created_at: &'a str,
        created_by: Option<&'a str>,
        last_updated_at: Option<&'a str>,
        object: T,
    ) -> Self {
        Record {
            id,
            created_at,
            created_by,
            last_updated_at,
            object,
        }
    }
}