【Rust问答】要如何实现一个全局变量的初始化（单例）

//创建连接
pub fn establish_connection() -> Arc {
    static mut POOL: Mutex>> = Mutex::new(None);
    unsafe {
        POOL.lock().unwrap().get_or_insert_with(||
            {
                println!("init pool ..");
                Arc::new(Pool::new(URL).unwrap())
            }
        )
            .clone()
    }
}

Mutex::new(None); static 变量中又不允许出现 非 const fn，怎么能实现这个需求呢

---

juzi5201314 2020-03-12 12:51

这个需求的话，可以看看lazy_static和once_cell这个库

juzi5201314 2020-03-12 12:58

once_cell:

use once_cell::sync::Lazy;

static POOL: Lazy = Lazy::new(|| Pool::new(URL).unwrap());
...
POOL.xxxx;


use once_cell::sync::OnceCell;

static POOL: OnceCell= OnceCell::new();
...
let pool = POOL.get_or_init(|| Pool::new(URL).unwrap());
...
POOL.set(Pool::new(URL).unwrap());
let pool = POOL.get().unwrap();

lazy_static:

lazy_static!{
static ref POOL: Mutex = Mutex::new(Pool::new(URL).unwrap());
}

juzi5201314 2020-03-12 13:09

嗷打少了Lazy,OnceCell,Mutex的 

phper-chen 2020-03-12 13:25

???? 

作者 LayneYy 2020-03-12 13:57

谢谢啦，刚接触rust，还不太熟练这个风格的变成，我发现很多都需要依赖标准库之外的

alexlee85 2020-03-18 17:53

如果你初始化之后不更新你的单例变量的话没必要用Mutex吧

use std::sync::Arc;

pub struct Pool {
    pub name: String,
}

pub fn establish_connection() -> Arc {
    static mut POOL: Option> = None;
    unsafe {
        Arc::clone(POOL.get_or_insert_with(|| {
            println!("init pool ~~~~~~~~~~~~");
            Arc::new(Pool {
                name: "I'm a pool".to_string(),
            })
        }))
    }
}

fn main() {
    let a = establish_connection();
    println!("{}", a.name);
    
    let b = establish_connection();
    println!("{}", b.name);
}

https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=3cf7b2701b7901e3db01838f9dec26ba