generics - 是否可以创建一个类型别名，该别名在函数的泛型类型上具有特征界限？

Question

这段代码:

pub type Foo<T: Read> = fn(bar: T);

产量 error E0122 (在较新版本的 Rust 中，这只是一个警告):

An attempt was made to add a generic constraint to a type alias. This constraint is entirely ignored. For backwards compatibility, Rust still allows this with a warning. Consider the example below:

trait Foo {}

type MyType<R: Foo> = (R, ());

fn main() {
    let t: MyType<u32>;
}

We're able to declare a variable of type MyType<u32>, despite the fact that u32 does not implement Foo. As a result, one should avoid using generic constraints in concert with type aliases.

是否可以创建一个包含函数指针特征要求的类型别名？显然，编译器告诉我类型不行，但我不知道是否还有其他我没有想到的函数选项。

Answer 1

对于从 Rust 1.47.0 开始仍然对此感到好奇的人来说，它仍然是不可能的，但看起来你会收到一 strip 有描述和建议替代方案的不错的小警告消息。 例如

pub type PublishQueue<T: From<Message>> = (tokio::sync::mpsc::Sender<T>);

产量

note: `#[warn(type_alias_bounds)]` on by default
help: the bound will not be checked when the type alias is used, and should be removed
|
| pub type PublishQueue<T> = sync::mpsc::Sender<T>;
|