如何为实现了特征的现有类型的枚举在范围内实现特征?
我有这个:
extern crate pnet;
use pnet::packet::ipv4::Ipv4Packet;
use pnet::packet::ipv6::Ipv6Packet;
enum EthernetType {
IPv4,
ARP,
VLAN,
IPv6,
Unknown(u16),
}
enum IPPacket<'a> {
IPv4(Ipv4Packet<'a>),
IPv6(Ipv6Packet<'a>),
}
fn ip_decode(pkt: &[u8]) -> IPPacket {
let version = (pkt[0] & 0xf0) >> 4;
if version == 4 {
IPPacket::IPv4(Ipv4Packet::new(&pkt).unwrap())
} else {
IPPacket::IPv6(Ipv6Packet::new(&pkt).unwrap())
}
}
fn main() {
// Parse ethernet packet here...
// ...
let ip_packet = ip_decode(b"deadbeef");
println!("{:?}", ip_packet.payload());
}
编译器抱怨我没有为我的枚举实现Packet
trait:
error[E0599]: no method named `payload` found for type `IPPacket<'_>` in the current scope
--> src/main.rs:32:32
|
14 | enum IPPacket<'a> {
| ----------------- method `payload` not found for this
...
32 | println!("{:?}", ip_packet.payload());
| ^^^^^^^
|
= help: items from traits can only be used if the trait is implemented and in scope
= note: the following trait defines an item `payload`, perhaps you need to implement it:
candidate #1: `pnet::packet::Packet`
我认为Packet
特性是通过Ipv4Packet<'a>
和Ipv6Packet<'a>
衍生的吗?
答案 0 :(得分:2)
如何为现有类型的枚举实现范围内的特征
为枚举实现特征的方式与为结构实现特征的方式相同:
trait Noise {
fn noise(&self);
}
enum Foo {
Bar,
Baz,
}
impl Noise for Foo {
fn noise(&self) {
match self {
Foo::Bar => println!("bar bar bar"),
Foo::Baz => println!("baz baz baz"),
}
}
}
已实现特征的现有类型的枚举
我认为
Packet
特质将会得到
不是。这样做会阻止人们在需要时实现他们的 own 代码。它也不能在所有情况下都起作用,例如当一个变体未实现时。
trait Noise {
fn noise(&self);
}
struct Bar;
impl Noise for Bar {
fn noise(&self) {
println!("bar bar bar");
}
}
struct Baz;
impl Noise for Baz {
fn noise(&self) {
println!("baz baz baz");
}
}
enum Foo {
Bar(Bar),
Baz(Baz),
}
impl Noise for Foo {
fn noise(&self) {
match self {
Foo::Bar(bar) => bar.noise(),
Foo::Baz(baz) => baz.noise(),
}
}
}
从概念上讲,可以对该语言进行扩展以支持某些注释来执行此操作,但我从未听说有人提出过建议。您可以考虑创建一个RFC来添加它。
也许您可以回到教导您的信息源并从根本上纠正问题,以防止其他人以相同的方式感到困惑。
另请参阅: