rust - 是什么使 `impl Trait` 作为参数 "universal"和作为返回值 "existential"？

Question

我正在阅读 RFC on "expanding" impl Trait当我遇到以下情况时:

By contrast, a programmer who first learned: fn take_iter(t: impl
Iterator) and then tried: fn give_iter() -> impl Iterator would be successful, without any rigorous understanding that they just transitioned from a universal to an existential.

虽然我从逻辑角度理解普遍性与存在性，但为什么第一个高于普遍性而第二个高于存在性？

Answer 1

RFC 以多种方式多次定义术语:

between existential types (where the callee chooses the type) and universal types (where the caller chooses)

There's been a lot of discussion around universals vs. existentials (in today's Rust, generics vs impl Trait).

• Universal quantification, i.e. "for any type T", i.e. "caller chooses". This is how generics work today. When you write fn foo<T>(t: T), you're saying that the function will work for any choice of T, and leaving it to your caller to choose the T.

• Existential quantification, i.e. "for some type T", i.e. "callee chooses". This is how impl Trait works today (which is in return position only). When you write fn foo() -> impl Iterator, you're saying that the function will produce some type T that implements Iterator, but the caller is not allowed to assume anything else about that type.

长话短说:

• fn take_iter(t: impl Iterator) — 调用 take_iter 的人选择具体类型。该函数必须适用于实现该特征的整个“宇宙”类型。

• fn give_iter() -> impl Iterator — give_iter 的实现选择具体类型。有一些类型“存在”并实现了函数将返回的特征。