rust - 将 TOML 反序列化为具有值的枚举向量

Question

我正在尝试读取 TOML 文件以创建一个结构，该结构包含具有关联值的枚举向量。这是示例代码:

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

use std::fs::File;
use std::io::Read;

#[derive(Debug, Deserialize, PartialEq)]
struct Actor {
    name: String,
    actions: Vec<Actions>,
}

#[derive(Debug, Deserialize, PartialEq)]
enum Actions {
    Wait(usize),
    Move { x: usize, y: usize },
}

fn main() {
    let input_file = "./sample_actor.toml";
    let mut file = File::open(input_file).unwrap();
    let mut file_content = String::new();
    let _bytes_read = file.read_to_string(&mut file_content).unwrap();
    let actor: Actor = toml::from_str(&file_content).unwrap();
    println!("Read actor {:?}", actor);
}

Cargo.toml

[dependencies]
serde_derive = "1.0.10"
serde = "1.0.10"
toml = "0.4.2"

sample_actor.toml

name = "actor1"
actions = [move 3 4, wait 20, move 5 6]

我知道该文件看起来“错误”，但我不知道我应该如何在 TOML 文件中编写操作，以便 crate 能够将它们识别为具有 X 个值的枚举。

使用 cargo run 运行此示例时出现的错误如下:

thread 'main' panicked at 'called `Result::unwrap()` on an `Err` value: Error { inner: ErrorInner { kind: NumberInvalid, line: Some(1), col: 11, message: "", key: [] } }', /checkout/src/libcore/result.rs:906:4
note: Run with `RUST_BACKTRACE=1` for a backtrace.

我知道我可能需要为我的枚举实现 FromStr 以将字符串转换为我的枚举，并且我简要地知道可以实现自定义反序列化器以特定方式反序列化，但我'我不确定这些部分是如何组合在一起的。

这似乎是一个使用 serde_json 而不是 TOML 的等效示例(当然是使用 JSON 文件而不是 TOML)。

代码的 JSON 版本:

extern crate serde;
extern crate serde_json;

#[macro_use]
extern crate serde_derive;

use serde_json::Error;
use std::fs::File;
use std::io::Read;

#[derive(Debug, Serialize, Deserialize)]
enum Foo {
    bar(u32),
    baz { x: u32, y: u32 },
}

#[derive(Debug, Serialize, Deserialize)]
struct Address {
    street: String,
    city: String,
    nums: Vec<Foo>,
}

fn main() {
    /*
        let address = Address {
            street: "10 Downing Street".to_owned(),
            city: "London".to_owned(),
            nums : vec![Foo::bar(1), Foo::baz{x : 2, y : 3}],
        };

        // Serialize it to a JSON string.
        let j = serde_json::to_string(&address).unwrap();

        // Print, write to a file, or send to an HTTP server.
        println!("{}", j);
    */
    let input_file = "./sample_address.json";
    let mut file = File::open(input_file).unwrap();
    let mut file_content = String::new();
    let _bytes_read = file.read_to_string(&mut file_content).unwrap();
    let address: Address = serde_json::from_str(&file_content).unwrap();
    println!("{:?}", address);
}

本例中读取/写入的JSON数据为:

Address { street: "10 Downing Street", city: "London", nums: [bar(1), baz { x: 2, y: 3 }] }

也许 TOML crate 不能支持我的用例？

Answer 1

Serde 有 lots of options for serializing enums .一个适合你的案例:

use serde::{Deserialize, Serialize}; // 1.0.117
use toml; // 0.5.7

#[derive(Debug, Serialize, Deserialize, PartialEq)]
#[serde(tag = "type", content = "args")]
enum Actions {
    Wait(usize),
    Move { x: usize, y: usize },
}

fn main() {
    let a_wait = Actions::Wait(5);
    println!("{}", toml::to_string(&a_wait).unwrap());

    let a_move = Actions::Move { x: 1, y: 1 };
    println!("{}", toml::to_string(&a_move).unwrap());
}

type = "Wait"
args = 5

type = "Move"

[args]
x = 1
y = 1