我可以选择特征边界吗?

给定一个模拟条件概率分布的特征:

trait Distribution {
    type T;
    fn sample<U>(&self, x: U) -> Self::T;
}

我想为两个结构实现特征, ConditionalNormalMultivariateConditionalNormal 分别为标量和向量值分布建模 .

这样的实现看起来像这样:

struct ConditionalNormal;

impl Distribution for ConditionalNormal {
    type T = f64;

    fn sample<U>(&self, x: U) -> Self::T {
        0.0
    }
}

struct MultivariateConditionalNormal;

impl Distribution for MultivariateConditionalNormal {
    type T = f64;

    fn sample<U>(&self, x: U) -> Self::T {
        0.0 + x[0]
    }
}

playground

但是, MultivariateConditionalNormal 的实现无效,因为通用 x[0] 不可索引 . 如果我添加特征边界 std::ops::Index<usize> ConditionalNormal 的实现无效,因为标量 f64 不可索引 .

我听说过,例如 Sized trait通过 ?Sized 接受可选的特征边界;我可以做类似的事吗?有什么方法可以解决这个问题吗?

回答(2)

3 years ago

您可以将特征的定义更改为

trait Distribution<U> {
    type T;
    fn sample(&self, x: U) -> Self::T;
}

这允许您在具有不同特征边界的各种类型上实现它 .

impl<U> Distribution<U> for ConditionalNormal {
    // ...
}

impl<U> Distribution<U> for MultivariateConditionalNormal
where
    U: std::ops::Index<usize, Output = f64>,
{
    // ...
}

Playground

3 years ago

您可以添加新特征以指定 U 的功能:

trait Distribution {
    type T;
    fn sample<U>(&self, x: U) -> Self::T
    where
        U: Samplable;
}

struct ConditionalNormal;

impl Distribution for ConditionalNormal {
    type T = f64;

    fn sample<U>(&self, x: U) -> Self::T
    where
        U: Samplable,
    {
        0.0.value()
    }
}

struct MultivariateConditionalNormal;

impl Distribution for MultivariateConditionalNormal {
    type T = f64;

    fn sample<U>(&self, x: U) -> Self::T
    where
        U: Samplable,
    {
        0.0 + x.value()
    }
}

trait Samplable {
    fn value(&self) -> f64;
}

impl Samplable for f64 {
    fn value(&self) -> f64 {
        *self
    }
}

impl Samplable for Vec<f64> {
    fn value(&self) -> f64 {
        self[0]
    }
}

fn main() {}