我有以下特征:
trait MyTrait {
type A;
type B;
fn foo(a: Self::A) -> Self::B;
fn bar(&self);
}
还有其他函数,如bar
,必须始终由特征的用户实现。
我想给foo
一个默认实现,但只有当类型A = B
.
伪锈代码:
impl??? MyTrait where Self::A = Self::B ??? {
fn foo(a: Self::A) -> Self::B {
a
}
}
这是可能的:
struct S1 {}
impl MyTrait for S1 {
type A = u32;
type B = f32;
// `A` is different from `B`, so I have to implement `foo`
fn foo(a: u32) -> f32 {
a as f32
}
fn bar(&self) {
println!("S1::bar");
}
}
struct S2 {}
impl MyTrait for S2 {
type A = u32;
type B = u32;
// `A` is the same as `B`, so I don't have to implement `foo`,
// it uses the default impl
fn bar(&self) {
println!("S2::bar");
}
}
这在 Rust 中可能吗?
您可以通过引入冗余类型参数在特征定义本身中提供默认实现:
trait MyTrait {
type A;
type B;
fn foo<T>(a: Self::A) -> Self::B
where
Self: MyTrait<A = T, B = T>,
{
a
}
}
对于单个类型,可以重写此默认实现。但是,专用版本将继承从特征上的foo()
定义绑定的 trait,因此只有在以下A == B
情况下才能实际调用该方法:
struct S1;
impl MyTrait for S1 {
type A = u32;
type B = f32;
fn foo<T>(a: Self::A) -> Self::B {
a as f32
}
}
struct S2;
impl MyTrait for S2 {
type A = u32;
type B = u32;
}
fn main() {
S1::foo(42); // Fails with compiler error
S2::foo(42); // Works fine
}
Rust 也有一个不稳定的 impl 专用化特性,但我认为它不能用来实现你想要的。
足够了吗?
trait MyTrait {
type A;
type B;
fn foo(a: Self::A) -> Self::B;
}
trait MyTraitId {
type AB;
}
impl<P> MyTrait for P
where
P: MyTraitId
{
type A = P::AB;
type B = P::AB;
fn foo(a: Self::A) -> Self::B {
a
}
}
struct S2;
impl MyTraitId for S2 {
type AB = i32;
}
锈游乐场
如前所述,如果MyTrait
MyTraitId
无法提供实现的其他方法,则会遇到问题。
扩展user31601的答案并使用Sven Marnach评论中的评论,以下是使用"委托方法"模式的附加函数实现的特征:
trait MyTrait {
type A;
type B;
fn foo(a: Self::A) -> Self::B;
fn bar();
}
trait MyTraitId {
type AB;
fn bar_delegate();
}
impl<P> MyTrait for P
where
P: MyTraitId,
{
type A = P::AB;
type B = P::AB;
fn foo(a: Self::A) -> Self::B {
a
}
fn bar() {
<Self as MyTraitId>::bar_delegate();
}
}
struct S2;
impl MyTraitId for S2 {
type AB = i32;
fn bar_delegate() {
println!("bar called");
}
}
fn main() {
<S2 as MyTrait>::bar(); // prints "bar called"
}
操场