How to preload lazy_static variables in Rust?

Question

I use Rust's lazy_static crate to assign a frequently used database object to a global variable but I do not want lazy loading. Is there a way to trigger lazy_static to preload the variable or is there a better way to achieve this? All the functions of the database are expensive and it seems to not be enough to assign a reference.

lazy_static! {
    static ref DB: DataBase = load_db();
}
 
/// this is too slow
#[allow(unused_must_use)] 
pub fn preload1() { DB.slow_function(); }

/// this does not cause lazy static to trigger
pub fn preload2() { let _ = &DB; }

Nit: Do not use lazy_static, use once_cell, its API is going to be integrated into std and now it is even recommended by the compiler. — Chayim Friedman
– Chayim Friedman, Commented Aug 31, 2022 at 22:04
@PitaJ: Yes, I forgot to change that from the real code to the example, thanks for noticing me! I fixed that. — Konrad Höffner
– Konrad Höffner, Commented Sep 1, 2022 at 7:06
@ChayimFriedman: I will do that! Interesting that the compiler recommends an external library, I have never seen that before in any language. — Konrad Höffner
– Konrad Höffner, Commented Sep 1, 2022 at 7:08
@KonradHöffner It was indeed doubted whether this should be implemented, but was now mainly because it is going to be part of the standard library. — Chayim Friedman
– Chayim Friedman, Commented Sep 1, 2022 at 7:15

PitaJ · Accepted Answer · 2022-08-31 16:44:57Z

1

_ does not bind, so use a variable name (with leading underscore to avoid the lint):

use lazy_static::lazy_static; // 1.4.0

struct DataBase;
fn load_db() -> DataBase {
    println!("db loaded");
    DataBase
}

lazy_static! {
    static ref DB: DataBase = load_db();
}

fn main() {
    let _db: &DataBase = &*DB; // prints "db loaded"
}

playground

answered Aug 31, 2022 at 16:44

PitaJ

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1 Comment

Chayim Friedman Over a year ago

The problem is not the _, it's the & that should be &*. Another solution is to type annotate as &DataBase to force a deref coercion, even with _.

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