How to serialize a u32 into byte array?

1

I am trying to serialize a u32 number into byte array, but I was not able until now, can someone help me with that?

Thanks

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2

You probably want byteorder. There are examples in the docs.

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3

If you don't want to depend on external libraries:

use std::mem::transmute;
let bytes: [u8; 4] = unsafe { transmute(123u32.to_be()) }; // or .to_le()

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4

Thank you guys, both solutions worked for me.

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5

If don't mind using operators and prefer not to use unsafe, you can do this

fn transform_u32_to_array_of_u8(x:u32) -> [u8;4] {
    let b1 : u8 = ((x >> 24) & 0xff) as u8;
    let b2 : u8 = ((x >> 16) & 0xff) as u8;
    let b3 : u8 = ((x >> 8) & 0xff) as u8;
    let b4 : u8 = (x & 0xff) as u8;
    return [b1, b2, b3, b4]
}
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6

This thread pops in search engines as one of the first hits, so the following might be useful to someone landing here.

From Rust 1.32.0 onward, the integer primitives have a builtin method for converting to a byte array:

fn main() {
    let value: u32 = 0x1FFFF;
    let bytes = value.to_be_bytes();
    println!("{:?}" , bytes);
}

Use to_le_bytes for Little Endian byte order.

No unsafe code and no external libraries needed.

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7

They also have from_{be,le}_bytes() as well-- nice!

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