如果我定义了一些枚举并且想要从字符串到该类型创建一个解析器,那么有什么比仅仅更好:
impl TheType {
fn from_str(s: &str) -> TheType {
// ...
}
}
从字符串/解析文本转换的正确方法是实现FromStr
特征。对于问题中的示例,它看起来像这样:
use std::str::FromStr;
enum Failure {
ReasonOne,
ReasonTwo,
}
impl FromStr for TheType {
type Err = Failure;
fn from_str(s: &str) -> Result<TheType, Self::Err> {
unimplemented!()
}
}
对于不能失败的泛型转换,您应该实现std::convert::From
特征:
use std::convert::From;
#[derive(PartialEq, Eq, Debug)]
enum MyEnum {
One,
Two,
Many(i64),
}
impl From<i64> for MyEnum {
fn from(val: i64) -> Self {
match val {
1 => MyEnum::One,
2 => MyEnum::Two,
_ => MyEnum::Many(val),
}
}
}
fn main() {
assert_eq!(MyEnum::from(1), MyEnum::One);
assert_eq!(MyEnum::from(2), MyEnum::Two);
assert_eq!(MyEnum::from(3), MyEnum::Many(3));
}
方便地,实现From
也自动实现Into
:
let one: MyEnum = 1.into(); assert_eq!(one, MyEnum::One);
let two: MyEnum = 2.into(); assert_eq!(two, MyEnum::Two);
let many: MyEnum = 3.into(); assert_eq!(many, MyEnum::Many(3));
对于可能失败的转换,您应该实现std::convert::TryFrom
。它仅在Rust 1.34及更高版本中可用,在这些版本之前,您可以使用conv
crate中的实现。