我有一个结构MyStruct
,需要一个通用的参数T: SomeTrait
,我想实施new
一个MyStruct
方法。这工作:
/// 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() {}
我希望把new
功能像这样的impl
块中:
impl MyStruct {
fn new<T: SomeTrait>(t: T) -> MyStruct<T> {
MyStruct { value: t }
}
}
但是,失败,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
如果我试图把它就像这样:
impl MyStruct<T> {
fn new(t: T) -> MyStruct<T> {
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
如何提供一个通用结构的实现?我应该把通用参数及其制约因素?
在<T: SomeTrait>
关键字之后的类型参数impl
应该来:
impl<T: SomeTrait> MyStruct<T> {
fn new(t: T) -> Self {
MyStruct { value: t }
}
}
如果类型和约束在impl<...>
列表变得太长,你可以使用where
语法和单独列出的限制:
impl<T> MyStruct<T>
where
T: SomeTrait,
{
fn new(t: T) -> Self {
MyStruct { value: t }
}
}
注意Self
的使用,这是可用的MyStruct<T>
块内impl
的快捷方式。
备注
impl<T>
的原因是this answer解释。本质上,它归结为一个事实,即impl<T> MyStruct<T>
和impl MyStruct<T>
是有效的,但意味着不同的事情。new
到impl
块,你应该删除多余的类型参数,否则你的结构的界面将变得不可用,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.
}