有没有一种办法不必两次初始化数组？ [英] Is there a way to not have to initialize arrays twice?

问题描述

我需要一个阵列的每个元素初始化为一个非恒定的恩pression。我可以这样做，而不必到阵列中的每个元素首先初始化一些相同的意义EX pression？下面是想我是能够做一个例子：

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 same meaningless expression? Here's an example of what I'd like to be able to do:

fn foo(xs: &[int, ..1000]) {
    let mut ys: [int, ..1000];

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

上面的code给出了编译时错误：结果
使用可能未初始化的变量的：伊苏

要解决这个问题，我需要改变功能的第一行，象这样一些虚拟值来初始化 YS 的元素：

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: [int, ..1000] = [42, ..1000];

有没有办法省略掉额外的初始化？包裹在一个不安全块一切似乎并没有任何区别。

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

推荐答案

使用的std ::纪念品::初始化，即

let mut ys: [int, ..1000] = unsafe{std::mem::uninitialized()};

我不认为这是一个更安全的方式来做到这一点。问题是，固定大小的数组只是没有得到同样的爱为 VEC 拉斯特，因为他们不能实施 FromIterator 。

I don't think there is a 'safer' way to do this. The problem is that constant size arrays just don't get the same love as Vec in Rust, as they can't implement FromIterator.

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