我想知道当两个或多个不同的枚举类型具有相同的数据成员或相同的函数时,是否有一种方法可以简化以下模式匹配臂。
(如果没有,最好解释一下原因)
UPDATE:
根据要求提供我想要的更准确的示例(请原谅我将数据成员访问与函数混淆)():
struct Point<T> {
x: i32,
y: T,
}
enum Record {
V4(Point<i64>),
V6(Point<i32>),
}
fn get_record() -> Record {
Record::V4(Point{ x: 1, y: 1})
}
fn main() {
let x = match get_record() {
Record::V4(r) => r.x,
Record::V6(r) => r.x,
};
println!("{}", &x);
// this will not compile
// let rec = get_record();
// println!("{}", rec.x);
// this will not compile either
// note: if V4 Point was i32 it will compile & run
// let rec = get_record();
// let x = match get_record() {
// Record::V4(r) | Record::V6(r) => r.x,
// };
}
原帖:
use std::net::IpAddr;
use std::str::FromStr;
fn main() {
let v4_or_v6 = IpAddr::from_str("1.2.3.4").unwrap();
// match expression, both arms only differ by 1 char
let s = match v4_or_v6 {
IpAddr::V4(ip) => ip.to_string(),
IpAddr::V6(ip) => ip.to_string(),
};
println!("{}", &s);
// not working:
// let s2 = match v4_or_v6 {
// IpAddr::V4(ip) | IpAddr::V6(ip) => ip.to_string(),
// };
// println!("{}", &s2);
}
我了解底层的调用to_string()
有不同的实现Ipv4
than Ipv6
但我认为编译器可以足够聪明来处理这个问题(我错了吗?)
尝试使用注释掉的代码进行编译会导致编译错误():
Compiling playground v0.0.1 (/playground)
error[E0308]: mismatched types
--> src/main.rs:16:37
|
16 | IpAddr::V4(ip) | IpAddr::V6(ip) => ip.to_string(),
| ^^ expected struct `std::net::Ipv4Addr`, found struct `std::net::Ipv6Addr`
|
= note: expected type `std::net::Ipv4Addr`
found type `std::net::Ipv6Addr`
error: aborting due to previous error
For more information about this error, try `rustc --explain E0308`.
error: Could not compile `playground`.