如果T实现了某个trait，那么整个T类型的&T，也实现了该trait了吗？不是这样的。
为什么这样，例子说明下：
trait Speaker {
fn speak(&self);
}
struct BasicSpeaker;
impl Speaker for BasicSpeaker {
fn speak(&self) {
println!("Hello!");
}
}
let speaker = BasicSpeaker;
speaker.speak();
结果可以正确运行，结果打印"Hello!"
如果实现对BasicSpeaker的引用，仍然调用speak，Rust的点号操作会自动解引用变量，所以代码仍然可以工作。
// 为了具体说明，具体给出类型
let speaker_ref: &BasicSpeaker = &speaker;
speaker_ref.speak();
这里不要认为rust会将T自动实现&T的trait。可以用函数参数方式传递，不会发生自动解引用时候。
fn speak_to(s: impl Speaker) {
s.speak();
}
fn main() {
// Create a BasicSpeaker struct:
let speaker = BasicSpeaker;
// Pass speaker to the new function:
speak_to(speaker);
}
直接调用没有问题，但是如果实现引用，就会出现问题:
let speaker_ref: &BasicSpeaker = &speaker;
speak_to(speaker_ref);
// 编译报错
error[E0277]: the trait bound `&BasicSpeaker: Speaker` is not satisfied
--> src/main.rs:31:14
|
31 | speak_to(speaker_ref);
| -------- ^^^^^^^^^^^ the trait `Speaker` is not implemented for `&BasicSpeaker`
| |
| required by a bound introduced by this call
|
= help: the trait `Speaker` is implemented for `BasicSpeaker`
note: required by a bound in `speak_to`
--> src/main.rs:16:21
|
16 | fn speak_to(s: impl Speaker) {
| ^^^^^^^ required by this bound in `speak_to`
For more information about this error, try `rustc --explain E0277`.
- 错误信息提示需要实现对&BasicSpeaker的trait Speaker才行。