如何将字节片(& [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，并且想要将其转换为整数:
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，但是在Rust中该怎么办?
I would do a cast in C, but what do I do in Rust?
推荐答案
我建议使用字节顺序箱(也可在无std环境中使用):
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);
}
这可以处理奇数大小的切片，并自动使缓冲区前进，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);
}
这些方法仅处理定长数组，以避免在没有足够数据时处理错误.如果有切片，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.
另请参阅:
- How to get a slice as an array in Rust?
- How to convert a slice into an array reference?
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