如何将GenericArray< T，?>成相同长度的数组? [英] How can I turn a GenericArray<T, ?> into an array of the same length?

问题描述

我正在计算给定数据的SHA256:

I'm computing the SHA256 of a given data:

let hashvalue = sha2::Sha256::digest(&data);

计算完后，我想将此值放入结构体的字段中:

After computing it, I want to put this value into a field of my struct:

let x = Hash { value: hashvalue };

但是， Hash 结构期望值的类型为 [u8;32] ，而我的 hashvalue 变量的类型为 GenericArray< u8，?> .如何将 hashvalue 转换为正确的类型?我试图将用作[u8;32] 和 arr！，但这没有用.

However, the Hash struct expects the type of value [u8; 32], while my hashvalue variable is of type GenericArray<u8, ?>. How can I convert hashvalue into the correct type? I tried to use as [u8; 32] and arr! but it didn't work.

推荐答案

如果您不知道数组的长度，请将 GenericArray 转换为切片，然后将切片转换为数组(仅适用于长度小于等于32的数组在Rust 1.47之前):

If you don't know the length of the array, convert the GenericArray into a slice and then convert the slice into an array (only for arrays of length 32 or less before Rust 1.47):

use sha2::Digest; // 0.9.3
use std::convert::TryInto;

fn main() {
    let hashvalue = sha2::Sha256::digest(&[3, 2, 6, 4, 3]);
    let x: [u8; 32] = hashvalue.as_slice().try_into().expect("Wrong length");
    println!("{:?}", x);
}

另请参阅:

• 如何在Rust中获取切片作为数组?
• 如何将切片转换为数组引用?
• 是否可以使用泛型的类型参数来控制数组的大小?

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