是否可以为由所有实现特征的类型组成的任何元组自动实现特征? [英] Is it possible to automatically implement a trait for any tuple that is made up of types that all implement the trait?
问题描述
假设我有一个
trait Happy {}
我可以为我想要的任何结构实现 Happy，例如:
I can implement Happy for whatever struct I might want, for example:
struct Dog;
struct Cat;
struct Alligator;
impl Happy for Dog {}
impl Happy for Cat {}
impl Happy for Alligator {}
现在，我想自动 impl 我的 Happy trait 来处理任何由实现 Happy trait 的类型组成的元组.直觉上，所有快乐的元组也是快乐的.
Now, I would like to automatically impl my Happy trait for whatever tuple is made up of types that all implement the Happy trait. Intuitively, a tuple of all happy is happy as well.
有没有可能做这样的事情?例如，我可以轻松地将 Happy 的实现扩展到两个 Happy 类型的任何元组:
Is it possible to do such a thing? For example, I can trivially extend the implementation of Happy to whatever tuple of two Happy types:
impl <T, Q> Happy for (T, Q) where T: Happy, Q: Happy {}
因此，这可以完美编译:
As a result, this compiles perfectly:
fn f(_: impl Happy) {
}
fn main() {
f((Dog{}, Alligator{}));
}
但是我如何将其推广到任何长度的元组?就我的理解而言，我们在 Rust 中没有可变泛型.有解决方法吗?
But how could I generalize that to any tuple, of any length? As far as my understanding goes, we don't have variadic generics in Rust. Is there a workaround?
推荐答案
我们在 Rust 中没有可变泛型.
we don't have variadic generics in Rust.
正确.
有解决方法吗?
您使用宏:
trait Happy {}
macro_rules! tuple_impls {
( $head:ident, $( $tail:ident, )* ) => {
impl<$head, $( $tail ),*> Happy for ($head, $( $tail ),*)
where
$head: Happy,
$( $tail: Happy ),*
{
// interesting delegation here, as needed
}
tuple_impls!($( $tail, )*);
};
() => {};
}
tuple_impls!(A, B, C, D, E, F, G, H, I, J,);
现在编译:
fn example<T: Happy>() {}
fn call<A: Happy, B: Happy>() {
example::<(A, B)>();
}
这通常不被视为大问题，因为长元组基本上是不可读的，如果确实需要，您总是可以嵌套元组.
This isn't generally seen as a big problem because long tuples are basically unreadable and you can always nest tuples if really needed.
另见:
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