包装类型时“在当前范围内找不到类型 T 的方法”

Question

我正在尝试围绕两种不同类型制作适配器来完成相同的工作，但我无法重写这两种类型。

X有一个消耗的方法self，因此运行时多态包装器不适用。唯一的选择是静态通用方法。

struct X {}

impl X {
    fn f(self, n: i32) {
        println!("n = {}", n);
    }
    fn new() -> X {
        X {}
    }
}

struct AdapterX {
    x: X
}

impl AdapterX {
    fn uf(self, n: i32) {
        self.x.f(n)
    }
    fn new(x: X) -> AdapterX {
        AdapterX { x: x }
    }
}

fn useAdapted<T>(a: T) {
    a.uf(10)
}

fn main() {
    let x = X::new();
    useAdapted::<AdapterX>(AdapterX::new(x));
}

编译器失败并显示：

struct X {}

impl X {
    fn f(self, n: i32) {
        println!("n = {}", n);
    }
    fn new() -> X {
        X {}
    }
}

struct AdapterX {
    x: X
}

impl AdapterX {
    fn uf(self, n: i32) {
        self.x.f(n)
    }
    fn new(x: X) -> AdapterX {
        AdapterX { x: x }
    }
}

fn useAdapted<T>(a: T) {
    a.uf(10)
}

fn main() {
    let x = X::new();
    useAdapted::<AdapterX>(AdapterX::new(x));
}

Answer 1

我能够弄清楚；不需要包装结构。正确的方法是一种普遍特征。我还错过了泛型类型变量的类型范围。

struct X {}

impl X {
    fn f(self, n: i32) {
        println!("n = {}", n);
    }
    fn new() -> X {
        X {}
    }
}

trait Adapter<T> {
    fn uf(self, n: i32);
}

impl Adapter<X> for X {
    fn uf(self, n: i32) {
        self.f(n)
    }
}

struct Y {}

impl Y {
    fn g(self, n: f32) {
        println!("m = {}", n);
    }
    fn new() -> Y {
        Y {}
    }
}

impl Adapter<Y> for Y {
    fn uf(self, n: i32) {
        self.g(n as f32)
    }
}

fn use_adapted<A, T: Adapter<A>>(a: T) {
    a.uf(10)
}

fn main() {
    use_adapted(X::new());
    use_adapted(Y::new());
}