Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
663 views
in Technique[技术] by (71.8m points)

initialization - Is there a way to not have to initialize arrays twice?

I need to initialize each element of an array to a non-constant expression. Can I do that without having to first initialize each element of the array to some meaningless expression? Here's an example of what I'd like to be able to do:

fn foo(xs: &[i32; 1000]) {
    let mut ys: [i32; 1000];

    for (x, y) in xs.iter().zip(ys.iter_mut()) {
        *y = *x / 3;
    }
    // ...
}

This code gives the compile-time error:

error[E0381]: borrow of possibly uninitialized variable: `ys`
 --> src/lib.rs:4:33
  |
4 |     for (x, y) in xs.iter().zip(ys.iter_mut()) {
  |                                 ^^ use of possibly uninitialized `ys`

To fix the problem, I need to change the first line of the function to initialize the elements of ys with some dummy values like so:

let mut ys: [i32; 1000] = [0; 1000];

Is there any way to omit that extra initialization? Wrapping everything in an unsafe block doesn't seem to make any difference.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

In some cases, you can use std::mem::MaybeUninit:

use std::mem::MaybeUninit;

fn main() {
    let mut ys: MaybeUninit<[i32; 1000]> = MaybeUninit::uninit();
}

Removing the MaybeUninit wrapper via assume_init is unsafe because accessing uninitialized values is undefined behavior in Rust and the compiler can no longer guarantee that every value of ys will be initialized before it is read.

Your specific case is one of the examples in the MaybeUninit docs; read it for a discussion about the safety of this implementation:

use std::mem::{self, MaybeUninit};

fn foo(xs: &[i32; 1000]) {
    // I copied this code from Stack Overflow without
    // reading why it is or is not safe.
    let ys: [i32; 1000] = {
        let mut ys: [MaybeUninit<i32>; 1000] = unsafe { MaybeUninit::uninit().assume_init() };

        let mut xs = xs.into_iter();

        for y in &mut ys[..] {
            if let Some(x) = xs.next().copied() {
                *y = MaybeUninit::new(x / 3);
            }
        }

        unsafe { mem::transmute(ys) }
    };
    // ...
}

You cannot collect into an array, but if you had a Vec instead, you could do:

let ys: Vec<_> = xs.iter().map(|&x| x / 3).collect();

For your specific problem, you could also clone the incoming array and then mutate it:

let mut ys = xs.clone();
for y in ys.iter_mut() { *y = *y / 3 }

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...