rust - 如何在没有结构实例的情况下调用特征方法？

Question

如果我有一个结构，其方法没有 self 作为参数，我可以通过 SomeStruct::method() 调用该方法。我似乎无法对从特征定义的方法做同样的事情。例如:

trait SomeTrait {
    fn one_trait() -> uint;
}

struct SomeStruct;
impl SomeStruct {
    fn one_notrait() -> uint {
        1u
    }
}
impl SomeTrait for SomeStruct {
    fn one_trait() -> uint {
        1u
    }
}

#[test]
fn testing() {
    SomeStruct::one_trait();   // doesn't compile
    SomeStruct::one_notrait(); // compiles
}

编译器给出错误“未解析的名称‘SomeStruct::one_trait.’”

如何直接调用结构的特征方法实现？

Answer 1

trait Animal {
    fn baby_name() -> String;
}

struct Dog;

impl Dog {
    fn baby_name() -> String {
        String::from("Spot")
    }
}

impl Animal for Dog {
    fn baby_name() -> String {
        String::from("puppy")
    }
}

fn main() {
    println!("A baby dog is called a {}", <Dog as Animal>::baby_name());
}

来自 Advanced Trait