是否可以为函数的泛型类型创建一个具有特征边界的类型别名? [英] Is it possible to create a type alias that has trait bounds on a generic type for a function?

问题描述

此代码:

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

产生 错误 E0122(在较新版本的Rust，这只是一个警告):

yields error E0122 (in newer versions of Rust, it is only a warning):

已尝试向类型别名添加通用约束.这个约束完全被忽略.为了向后兼容，Rust仍然允许此警告.考虑以下示例:

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>;
}

我们可以声明一个 MyType 类型的变量，尽管事实上u32 没有实现 Foo.因此，应避免使用与类型别名一致的通用约束.

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对函数指针的特征要求的类型别名?显然，编译器告诉我没有类型，但不知道是否还有其他我没有想到的函数选项.

Is it possible to create a type alias that contains trait requirements on a function pointer? Obviously the compiler is telling me no for types, but didn't know if there was another option for functions that I wasn't thinking of.

推荐答案

目前看来不可能，也没有解决方法.

At this time, it does not seem to be possible and no workarounds exist.

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