如何将一段字节 (&[u8]) 的缓冲区转换为整数? [英] How can I convert a buffer of a slice of bytes (&[u8]) to an integer?

问题描述

我正在从文件中读取原始数据，我想将其转换为整数:

I am reading raw data from a file and I want to convert it to an integer:

fn main() {
    let buf: &[u8] = &[0, 0, 0, 1];
    let num = slice_to_i8(buf);
    println!("1 == {}", num);
}

pub fn slice_to_i8(buf: &[u8]) -> i32 {
    unimplemented!("what should I do here?")
}

我会在 C 中进行强制转换，但我会在 Rust 中做什么?

I would do a cast in C, but what do I do in Rust?

推荐答案

我建议使用 byteorder crate(也适用于非标准环境):

I'd suggest using the byteorder crate (which also works in a no-std environment):

use byteorder::{BigEndian, ReadBytesExt}; // 1.2.7

fn main() {
    let mut buf: &[u8] = &[0, 0, 0, 1];
    let num = buf.read_u32::<BigEndian>().unwrap();

    assert_eq!(1, num);
}

这会处理奇怪大小的切片并自动推进缓冲区，以便您可以读取多个值.

This handles oddly-sized slices and automatically advances the buffer so you can read multiple values.

从 Rust 1.32 开始，您还可以使用 from_le_bytes/from_be_bytes/from_ne_bytes 整数的固有方法:

As of Rust 1.32, you can also use the from_le_bytes / from_be_bytes / from_ne_bytes inherent methods on integers:

fn main() {
    let buf = [0, 0, 0, 1];
    let num = u32::from_be_bytes(buf);

    assert_eq!(1, num);
}

这些方法只处理固定长度的数组，以避免在数据不足时处理错误.如果您有切片，则需要将其转换为数组.

These methods only handle fixed-length arrays to avoid dealing with the error when not enough data is present. If you have a slice, you will need to convert it into an array.

另见:

• 如何在 Rust 中将切片作为数组获取?
• 如何将切片转换为数组引用?

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