rust - 如何在Rust中为返回的元组设置静态生存期？

Question

假设我有以下返回元组的构造函数:

pub struct WebCam {
    id: u8
}

impl WebCam {
    fn new() -> (Self, bool) {
        let w = WebCam {id: 1 as u8};
        return (w, false);
    }
}

pub fn main() {
    static (cam, isRunning): (WebCam, bool) = WebCam::new();
}

上面的代码无法编译。但是，如果我将static更改为let，则可以正常编译。我不确定如何在返回的元组的单个值上设置生命周期(一行)？

Answer 1

https://doc.rust-lang.org/reference/items/static-items.html

是否应该使用常量项目还是静态项目可能会造成混淆。通常，常量应优先于静态，除非满足以下条件之一:

正在存储大量数据
静态变量的单地址属性是必需的。
内部可变性是必需的。

A static item is similar to a constant, except that it represents a precise memory location in the program. All references to the static refer to the same memory location. Static items have the static lifetime, which outlives all other lifetimes in a Rust program. Non-mut static items that contain a type that is not interior mutable may be placed in read-only memory. Static items do not call drop at the end of the program.

All access to a static is safe, but there are a number of restrictions on statics:

The type must have the Sync trait bound to allow thread-safe access. Statics allow using paths to statics in the constant expression used to initialize them, but statics may not refer to other statics by value, only through a reference. Constants cannot refer to statics.

我会像这样重写您的代码:

pub struct WebCam {
    id: u8,
}

impl WebCam {
    fn new() -> (Self, bool) {
        (WebCam { id: 1u8 }, false)
    }
}

pub fn main() {
    let (cam, is_running) = WebCam::new();
    println!("{} {}", cam.id, is_running);
}

这也适用:

pub struct WebCam {
    id: u8,
}

impl WebCam {
    fn new() -> (Self, bool) {
        (WebCam { id: 1u8 }, false)
    }
}
static mut IS_RUNNING: bool = false;
static mut WEB_CAM: WebCam = WebCam { id: 0u8 };

pub fn main() {
    let (cam, is_running) = WebCam::new();

    unsafe {
        IS_RUNNING = is_running;
        WEB_CAM.id = cam.id;
    }

    println!("{} {}", cam.id, is_running);
}