我如何定义一个返回自己类型的Rust函数类型? [英] How can I define a Rust function type which returns its own type?
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问题描述
$ b
type FnType func(paramType)FnType
code>
它只是一个返回相同类型函数的函数。 Rust中可以实现类似的东西吗?而且,理想情况下,是否可以一般地完成,以便客户端指定 paramType
?
解决方案
我在文档中进行了一些挖掘工作,并将它们带到了操场上,我认为我自己能够回答这个问题,尽管它需要一个中间类型: enum <
fn main(){
let mut state = State :: code>一些(第一);
,同时让State :: Some(s)= state {
state = s(0)
}
}
enum State< T> {
一些(fn(T) - > State< T>),
None,
}
fn first(_:i32) - >状态< I32> {
println!(First);
State :: Some(second)
}
fn second(_:i32) - >状态< I32> {
println!(Second);
State :: None
}
您可以验证它是否在游乐场。
I'm learning Rust, and still very much trying to get my head around it. Consider the following Go definition:
type FnType func(paramType) FnType
It's just a function that returns a function of the same type. Can something similar be implemented in Rust? And, ideally, can it be done generically, so that paramType
is specified by the client?
解决方案
I did some digging in the docs and took to the playground and I think I've been able to answer this myself, although it does require an intermediary type: an enum
, to be specific.
fn main() {
let mut state = State::Some(first);
while let State::Some(s) = state {
state = s(0)
}
}
enum State<T> {
Some(fn(T) -> State<T>),
None,
}
fn first(_: i32) -> State<i32> {
println!("First");
State::Some(second)
}
fn second(_: i32) -> State<i32> {
println!("Second");
State::None
}
You can verify that it runs on the playground.
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