如何在 structopt 中使用枚举？

Question

我想让 StructOpt 使用枚举，这样每次用户通过 -d sunday 它都会被解析为 Day::Sunday:

#[macro_use]
extern crate structopt;

use std::path::PathBuf;
use structopt::StructOpt;

// My enum
enum Day {
    Sunday, Monday
}

#[derive(Debug, StructOpt)]
#[structopt(name = "example", about = "An example of StructOpt usage.")]
struct Opt {
    /// Set speed
    #[structopt(short = "s", long = "speed", default_value = "42")]
    speed: f64,
    /// Input file
    #[structopt(parse(from_os_str))]
    input: PathBuf,
    /// Day of the week
    #[structopt(short = "d", long = "day", default_value = Day::Monday)]
    day: Day,
}

fn main() {
    let opt = Opt::from_args();
    println!("{:?}", opt);
}

我目前的最佳解决方案是使用Option<String>作为类型并传递自定义parse_day():

fn parse_day(day: &str) -> Result<Day, ParseError> {
    match day {
        "sunday" => Ok(Day::Sunday),
        _ => Ok(Day::Monday)
    }
    Err("Could not parse a day")
}

Answer 1

Struct-opt 接受任何实现了 FromStr 的类型，它离你的 parse_day 函数不远：

use std::str::FromStr;

// any error type implementing Display is acceptable.
type ParseError = &'static str;

impl FromStr for Day {
    type Err = ParseError;
    fn from_str(day: &str) -> Result<Self, Self::Err> {
        match day {
            "sunday" => Ok(Day::Sunday),
            "monday" => Ok(Day::Monday),
            _ => Err("Could not parse a day"),
        }
    }
}

此外，default_value 应该是一个字符串，它将使用 from_str.

解释为 Day

#[structopt(short = "d", long = "day", default_value = "monday")]
day: Day,

Answer 2

错误信息是：

error[E0277]: the trait bound `Day: std::str::FromStr` is not satisfied
  --> src/main.rs:22:17
   |
22 | #[derive(Debug, StructOpt)]
   |                 ^^^^^^^^^ the trait `std::str::FromStr` is not implemented for `Day`
   |
   = note: required by `std::str::FromStr::from_str`

您可以通过为 Day 实现 FromStr（参见）来解决该问题，正如消息所暗示的那样，或者通过为 Day 定义解析函数：

fn parse_day(src: &str) -> Result<Day, String> {
    match src {
        "sunday" => Ok(Day::Sunday),
        "monday" => Ok(Day::Monday),
        _ => Err(format!("Invalid day: {}", src))
    }
}

并使用 try_from_str 属性指定它：

/// Day of the week
#[structopt(short = "d", long = "day", parse(try_from_str = "parse_day"), default_value = "monday")]
day: Day,

Answer 3

works, but the arg_enum! macro is a more concise way of doing it, as demonstrated in this example from structopt:

arg_enum! {
    #[derive(Debug)]
    enum Day {
        Sunday,
        Monday
    }
}

#[derive(StructOpt, Debug)]
struct Opt {
    /// Important argument.
    #[structopt(possible_values = &Day::variants(), case_insensitive = true)]
    i: Day,
}

fn main() {
    let opt = Opt::from_args();
    println!("{:?}", opt);
}

这会让您将工作日解析为 Sunday 或 sunday。

Answer 4

显然 arg_enum! 已被弃用，clap 3 即将推出，see the doc here。我只是重构为直接使用clap而不是StructOpts，看来它们已经合并了。

在您的 Cargo.toml 中，您不再需要 StructOpt：

clap = { version = "3.1.17", features = ["derive"] }

在你的 main.rs 中，你现在将使用 clap 和内置的 struct opt，注意字段上的文档变成了整洁的字段帮助消息值：

use clap::StructOpt;

#[derive(clap::ArgEnum, Debug, Clone)]
enum PossibleWords {
    Hello,
    World,
}

#[derive(Debug, clap::Parser, Clone)]
#[clap(long_about = "The worst Hello World!!! App in the world!")]
pub struct Args {
    /// You can't have everything in life, which word should we print?
    #[clap(arg_enum, long, default_value_t = YourEnum::Hello)]
    word: PossibleWords,

    //// Exclamation marks are included FOR FREE with the word of your choice.
    #[clap(long, default_value_t = 10)]
    number_of_exclamation_marks: u64,
}

pub fn main() {
    let args = Args::parse();
}

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