我可以在 Rust 中将字符串转换为没有宏的枚举吗? [英] Can I convert a string to enum without macros in Rust?

问题描述

例如，如果我有这样的代码:

For example, if I have code like:

enum Foo {
    Bar,
    Baz,
    Bat,
    Quux
}

impl Foo {
    from(input: &str) -> Foo {
        Foo::input
    }
}

这显然会失败，因为 input 不是 Foo 的方法.我可以手动输入:

This will obviously fail because input is not a method of Foo. I can manually type:

from(input: &str) -> Foo {
    match(input) {
        "Bar" => Foo::Bar,
        // and so on...
    }
}

但我没有得到自动的便利.

but I'm not getting the automatic convenience.

看起来 Java 在枚举上有一个 字符串查找函数 用于此特定目的.

It looks like Java has a string lookup function on enums for this specific purpose.

是否有可能在不编写我自己的宏或从 crate 中导入宏的情况下获得它?

Is it possible to get this without writing my own macro or importing one from a crate?

推荐答案

你应该实现 std::str::FromStr 特征.

use std::str::FromStr;

#[derive(Debug, PartialEq)]
enum Foo {
    Bar,
    Baz,
    Bat,
    Quux,
}

impl FromStr for Foo {

    type Err = ();

    fn from_str(input: &str) -> Result<Foo, Self::Err> {
        match input {
            "Bar"  => Ok(Foo::Bar),
            "Baz"  => Ok(Foo::Baz),
            "Bat"  => Ok(Foo::Bat),
            "Quux" => Ok(Foo::Quux),
            _      => Err(()),
        }
    }
}



fn main() {
    // Use it like this
    let f = Foo::from_str("Baz").unwrap();
    assert_eq!(f, Foo::Baz);
}

代码生成(又称自动便利)和反射通常需要付出代价.在实践中，您最终不太可能得到多个枚举变体.

Code-generation (aka automatic convenience) and reflections usually bear a cost. In practice, it is unlikely that you will end up with more than a few enum variants.

在游乐场

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