Generics 如何在Rust中提供通用结构的实现?
我有一个结构Generics 如何在Rust中提供通用结构的实现?,generics,struct,rust,Generics,Struct,Rust,我有一个结构MyStruct,它接受一个通用参数T:SomeTrait,我想为MyStruct实现一个新的方法。这项工作: /// Constraint for the type parameter `T` in MyStruct pub trait SomeTrait: Clone {} /// The struct that I want to construct with `new` pub struct MyStruct<T: SomeTrait> { value:
MyStruct
,它接受一个通用参数T:SomeTrait
,我想为MyStruct
实现一个新的方法。这项工作:
/// Constraint for the type parameter `T` in MyStruct
pub trait SomeTrait: Clone {}
/// The struct that I want to construct with `new`
pub struct MyStruct<T: SomeTrait> {
value: T,
}
fn new<T: SomeTrait>(t: T) -> MyStruct<T> {
MyStruct { value: t }
}
fn main() {}
但这无法通过以下方式编译:
error[E0107]: wrong number of type arguments: expected 1, found 0
--> src/main.rs:9:6
|
9 | impl MyStruct {
| ^^^^^^^^ expected 1 type argument
如果我试着这样说:
impl MyStruct {
fn new<T: SomeTrait>(t: T) -> MyStruct<T> {
MyStruct { value: t }
}
}
impl MyStruct<T> {
fn new(t: T) -> MyStruct<T> {
MyStruct { value: t }
}
}
impl MyStruct{
fn new(t:t)->MyStruct{
MyStruct{value:t}
}
}
错误更改为:
error[E0412]: cannot find type `T` in this scope
--> src/main.rs:9:15
|
9 | impl MyStruct<T> {
| ^ not found in this scope
错误[E0412]:在此作用域中找不到类型'T'
-->src/main.rs:9:15
|
9 |实施我的结构{
|^在此范围内找不到
如何提供泛型结构的实现?将泛型参数及其约束放在何处?类型参数
应紧跟在impl
关键字之后:
impl<T: SomeTrait> MyStruct<T> {
fn new(t: T) -> Self {
MyStruct { value: t }
}
}
请注意Self
的用法,它是impl
块中可用的MyStruct
的快捷方式
备注
需要impl
的原因在中进行了解释。从本质上讲,这归结为impl MyStruct
和impl MyStruct
都是有效的,但含义不同
将new
移动到impl
块时,应删除多余的类型参数,否则结构的接口将变得不可用,如下例所示:
// trait SomeTrait and struct MyStruct as above
// [...]
impl<T> MyStruct<T>
where
T: SomeTrait,
{
fn new<S: SomeTrait>(t: S) -> MyStruct<S> {
MyStruct { value: t }
}
}
impl SomeTrait for u64 {}
impl SomeTrait for u128 {}
fn main() {
// just a demo of problematic code, don't do this!
let a: MyStruct<u128> = MyStruct::<u64>::new::<u128>(1234);
// ^
// |
// This is an irrelevant type
// that cannot be inferred. Not only will the compiler
// force you to provide an irrelevant type, it will also
// not prevent you from passing incoherent junk as type
// argument, as this example demonstrates. This happens
// because `S` and `T` are completely unrelated.
}
//trait SomeTrait和struct MyStruct如上所述
// [...]
impl MyStruct
哪里
T:某种特质,
{
fn新(t:S)->MyStruct{
MyStruct{value:t}
}
}
u64{}的impl SomeTrait
u128{}的impl SomeTrait
fn main(){
//只是一个有问题代码的演示,不要这样做!
设a:MyStruct=MyStruct:::new::(1234);
// ^
// |
//这是一种不相关的类型
//这是无法推断的。不仅编译器
//强制您提供一个不相关的类型,它也会
//不阻止您将不连贯的垃圾作为类型传递
//参数,如本例所示。这种情况会发生
//因为'S'和'T'完全不相关。
}
@AndreyTyukin我刚刚找到了你的答案,正在阅读std::rc::rc in的代码。如果你创建了答案,我会很乐意标记它。这不是一个拼写错误;我只是不知道“impl”会起作用。好的。我也投票关闭了。如果你在答案中添加链接可能是最好的,这样它就不会在这些短暂的评论中丢失。回答很好。比ks提供了详细的解释和清晰的示例。
// trait SomeTrait and struct MyStruct as above
// [...]
impl<T> MyStruct<T>
where
T: SomeTrait,
{
fn new<S: SomeTrait>(t: S) -> MyStruct<S> {
MyStruct { value: t }
}
}
impl SomeTrait for u64 {}
impl SomeTrait for u128 {}
fn main() {
// just a demo of problematic code, don't do this!
let a: MyStruct<u128> = MyStruct::<u64>::new::<u128>(1234);
// ^
// |
// This is an irrelevant type
// that cannot be inferred. Not only will the compiler
// force you to provide an irrelevant type, it will also
// not prevent you from passing incoherent junk as type
// argument, as this example demonstrates. This happens
// because `S` and `T` are completely unrelated.
}