返回具有关联类型的特征 [英] Returning a trait with an associated type

问题描述

struct A;
struct PropA;

struct B;
struct PropB;

trait AB{
    type prop;
    fn a(&self) -> ();
    fn b(&self, p: Self::prop) -> ();
}

impl AB for A{
    type prop = PropA;
    fn a(&self)->(){}
    fn b(&self, p: Self::prop) -> (){}
}
impl AB for B{
    type prop = PropB;
    fn a(&self)->(){}
    fn b(&self, p: Self::prop) -> (){}
}

fn get_a_or_b(s: &str) -> Option<Box<dyn AB<prop=_>>>{
    match s{
        "a" => Some(Box::new(A)),
        "b" => Some(Box::new(B)),
        _=> None
    }
}

游乐场链接

我将基于字符串输入返回两个不同的结构 A & B .

I am returning two different structs A&B based on a string input.

在将相关类型指定为占位符时，我得到在项目签名上的类型中不允许使用占位符'_'.

I get the type placeholder '_' is not allowed within types on item signatures when specifying associated type as placeholder.

推荐答案

我认为这里存在误解； dyn AB< Prop = A> 和 dyn AB< Prop = B> 是不同的类型，第一个是动态的 AB< Prop = A> >，第二个是动态的 AB< Prop = B> .这意味着您不能将泛型类型和关联类型留给动态方面.

I believe there's a misconception here; dyn AB<Prop = A> and dyn AB<Prop = B> are different types, the first is a dynamic AB<Prop = A> and the second is a dynamic AB<Prop = B>. What this means is that you cannot leave generic types and associated types up to the dynamic aspect.

这与未提及关联类型时不同:

This is different than when the associated type is not mentioned:

fn foo<T: AB>() {
    let my_fn: fn(&T, T::Prop) = T::b;
}

我们在这里访问 T :: Prop 而不是分配它.

Where we access the T::Prop instead of assigning it.

所有类型都必须是具体的，并且一个分支上的 dyn AB< Prop = A> 和另一分支上的 dyn AB< Prop = B> 并不具体，但是应该将它包装在枚举下:

All types must be concrete, and dyn AB<Prop = A> on one branch, and dyn AB<Prop = B> on another branch is not concrete, but could be should you package it up under an enum:

enum AOrB {
    A(Box<dyn AB<Prop = A>>),
    B(Box<dyn AB<Prop = B>>),
}

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