为什么在泛型函数中对特征的引用必须实现“Sized"? [英] Why does a reference to a trait in a generic function have to implement `Sized`?

问题描述

我有一个函数返回对 trait 的引用(trait_ref())和另一个函数，它接受对通用 trait 实现的引用(take_trait_ref_generic).

I have a function that returns a reference to a trait (trait_ref()) and another function that takes a reference to a generic trait implementation (take_trait_ref_generic).

但是，无法将我从第一个函数获得的引用传递给第二个函数.Rustc 抱怨特征 std::marker::Sized 没有为 SomeTrait 实现".

However, it's not possible to pass the reference I get from the first function to the second one. Rustc complains that "the trait std::marker::Sized is not implemented for SomeTrait".

即使是这样，为什么它必须实现Sized?反正都是参考.

Even though that's true, why does it have to implement Sized? It's a reference anyway.

trait SomeTrait {}

struct TraitImpl;

impl SomeTrait for TraitImpl {}

struct Container {
    trait_impl: TraitImpl,
}

impl Container {
    fn trait_ref(&self) -> &SomeTrait {
        &self.trait_impl
    }
}

fn take_trait_ref_generic<T: SomeTrait>(generic_trait_ref: &T) {}

fn main() {
    let container = Container { trait_impl: TraitImpl };

    /*Not possible*/
    take_trait_ref_generic(container.trait_ref());
}

推荐答案

默认情况下，所有 函数的泛型类型都隐式地具有 Sized 绑定，无论它们如何用过的.您需要使用 ?Sized 明确选择退出该要求:

By default, all generic types on functions implicitly have the Sized bound, regardless of how they are used. You need to explicitly opt-out of that requirement using ?Sized:

fn take_trait_ref_generic<T>(generic_trait_ref: &T)
where 
    T: ?Sized + SomeTrait
{}

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