如何表达某些返回类型实现特征？

Question

这是显示我的问题的抽象示例,从我尝试重构一些Rust代码并同时学习Rust中提取.

struct GenStruct<T> {
    field: T,
}

trait Marker {}
trait Return {}

impl Marker for i32 {}
impl Marker for u32 {}

// actually implement `Return` for GenStruct<M: Marker>,
// but compiler don't recognize that
impl Return for GenStruct<i32> {}
impl Return for GenStruct<u32> {}

struct Fake;

trait Trait<M: Marker> {
    type Ret: Return;
    fn meth(m: M) -> Self::Ret;
}

impl<M: Marker> Trait<M> for Fake {
    type Ret = GenStruct<M>;

    fn meth(m: M) -> GenStruct<M> {
        GenStruct { field: m }
    }
}

输出:

error[E0277]: the trait bound `GenStruct<M>: Return` is not satisfied
  --> src/lib.rs:23:17
   |
23 | impl<M: Marker> Trait<M> for Fake {
   |                 ^^^^^^^^ the trait `Return` is not implemented for `GenStruct<M>`
   |

编译器无法识别我实际Return为GenStruct<M>M所在的每个地方实现Marker.为了解决这个问题,我可以这样写:

trait Marker {
    fn is_i32() -> bool;
}
trait Return {
    fn ret();
}

impl Marker for i32 {
    fn is_i32() -> bool {
        true
    }
}
impl Marker for u32 {
    fn is_i32() -> bool {
        false
    }
}

// compiler is satisfied by such implementation
impl<M: Marker> Return for GenStruct<M> {
    fn ret() {
        if M::is_i32() {
        } else {
        }
    }
}

..或使用特质对象..

impl<M: Marker> Return for GenStruct<M> {}

trait Trait<'a, M: Marker + 'a> {
    fn meth(m: M) -> Box<Return + 'a>;
}

impl<'a, M: Marker + 'a> Trait<'a, M> for Fake {
    fn meth(m: M) -> Box<Return + 'a> {
        Box::new(GenStruct { field: m })
    }
}

..但如果我使用特征对象,我不能写一个专门的Returnfor GenStruct<i32>和GenStruct<u32>.

我的问题是:能否GenStruct<M: Marker>认识到我确实实现了Return或者我的代码是不是惯用的Rust？如果我的代码不是惯用的,那么编写代码的正确方法是什么？

Answer 1

使用where子句,我们可以在泛型上添加一个额外的约束,impl以便它只适用于GenStruct<M>确实实现的情况Return.

impl<M: Marker> Trait<M> for Fake
where
    GenStruct<M>: Return,
{
    type Ret = GenStruct<M>;

    fn meth(m: M) -> GenStruct<M> {
        GenStruct { field: m }
    }
}