防止特征函数被其他结构实现

Question

我只是构建了一个Bar具有 2 个函数的特征（alpha()带有实现和beta()仅带有接口），并且我希望实现的结构Bar只实现beta()，而不是实现自己的alpha()。

有什么方法可以阻止另一个结构实现自己的结构alpha()吗？

trait Bar {
    fn alpha(&self) {
        println!("you should never implement this function on your own.");
    }

    fn beta(&self);
}

struct Foo {}


impl Bar for Foo {
    fn alpha(&self) {
        println!("how do I trigger an error here when a struct implement it's own alpha()?");
    }

    fn beta(&self) {
        println!("implement beta() for Foo");
    }
}

Answer 1

您可以通过将您的特征分成两部分，并使用默认方法为该特征提供一揽子实现来做到这一点。

pub trait Beta {
    fn beta(&self);
}

pub trait Alpha: Beta {
    fn alpha(&self) {
        // Default impl
    }
}

// Because of this blanket implementation,
// no type can implement `Alpha` directly,
// since it would conflict with this impl.
// And you cannot implement `Alpha` without `Beta`,
// since `Beta` is its _supertrait_.
impl<T: Beta> Alpha for T {}


struct Foo;

impl Beta for Foo {
    fn beta(&self) {
        // Impl
    }
}

// If you uncomment this you will have a compile error,
// because of the conflicting implementations of Alpha for Foo
// (conflicts with the blanket implementation above)
//
// impl Alpha for Foo {
//    fn alpha(&self) {
//        // override impl
//    } 
// }

pub fn bar<T: Alpha>(_: T) {}

pub fn baz() {
    bar(Foo);
}