【发布时间】:2018-08-02 20:06:31
【问题描述】:
如何在已实现该特征的现有类型的枚举范围内实现特征?
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 trait 将通过Ipv4Packet<'a> 和Ipv6Packet<'a> 派生?
【问题讨论】:
-
在提供的代码示例中,您还没有实现任何特征。你忘了粘贴什么吗?