问题描述
我正在尝试实现一个包含通用方法的特征.
I'm trying to implement a trait which contains a generic method.
trait Trait {
fn method<T>(&self) -> T;
}
struct Struct;
impl Trait for Struct {
fn method(&self) -> u8 {
return 16u8;
}
}
我明白了:
error[E0049]: method `method` has 0 type parameters but its trait declaration has 1 type parameter
--> src/lib.rs:8:5
|
2 | fn method<T>(&self) -> T;
| ------------------------- expected 1 type parameter
...
8 | fn method(&self) -> u8 {
| ^^^^^^^^^^^^^^^^^^^^^^ found 0 type parameters
我应该如何正确编写 impl 块?
How should I write the impl block correctly?
推荐答案
函数和方法中的类型参数通用.这意味着对于所有 trait 实现者，Trait::method 必须为任何 T 实现，其约束与 trait 所指示的约束完全相同(在这种情况下，T 上的约束只是隐含的 Sized).
Type parameters in functions and methods are universal. This means that for all trait implementers, Trait::method<T> must be implemented for any T with the exact same constraints as those indicated by the trait (in this case, the constraint on T is only the implicit Sized).
您指出的编译器错误消息表明它仍然需要参数类型 T.相反，您的 Struct 实现假设 T = u8，这是不正确的.类型参数由方法的调用者而不是实现者决定，因此 T 可能并不总是 u8.
The compiler's error message that you indicated suggests that it was still expecting the parameter type T. Instead, your Struct implementation is assuming that T = u8, which is incorrect. The type parameter is decided by the caller of the method rather than the implementer, so T might not always be u8.
如果您希望让实现者选择特定类型，则必须改为在关联类型中具体化.
If you wish to let the implementer choose a specific type, that has to be materialized in an associated type instead.
trait Trait {
type Output;
fn method(&self) -> Self::Output;
}
struct Struct;
impl Trait for Struct {
type Output = u8;
fn method(&self) -> u8 {
16
}
}
另请阅读Rust 编程语言的这一部分:在具有关联类型的特征定义中指定占位符类型.
Read also this section of The Rust Programming Language: Specifying placeholder types in trait definitions with associated types.
另见:
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