Rust bincode enum binary serialisation

Question

I'm fairly new to Rust and am using bincode in my project. The encoding of enum variants is an issue that I've been trying to deal with as I am interfacing with an existing server. The old project was written in C and the compilier option of "fshort-enums" was defined.

In the old C project I would define an enum something like:

enum test_enum {
    Start,
    Init,
    Complete,
}

and then when this data was put into the packet it would take up 1 byte like this:

0x00,
0x01,
0x02

If I defined the enum as:

enum test_enum {
    Start,
    Init,
    Complete = 65535,
}

then when this data was put into the packet every variant would take up 2 bytes like this:

0x00 0x00,
0x01 0x00,
0x02 0x00

Now I've found a way to configure bincode enum encoding to change from the default 4 bytes to be the appropriate size for the value like this:

#[derive(Serialize, Deserialize)]
enum TestEnum {
    Start,
    Init,
    Complete = 65535,
}

let x = TestEnum::Complete;
if let Ok(bytes) = bincode::options().with_varint_encoding().serialize(&x) {
    println!("{:?}", bytes); // [2]
}

Question1: How do I force bincode to make all variants take up 2 bytes?

Question2: How do I force bincode to apply the enum variant value? (in this example TestEnum::Complete has a value of 2 instead of 65535).

the point of bincode is to handle this for you, you can't config these — Stargateur, Oct 24 '20 at 10:25

Peter Hall · Answer 1 · 2020-10-24T12:07:16.500

Options::with_varint_encoding is probably not what you want, since it will only use 1 byte for integers that are less than 256. You can use Options::with_fixint_encoding to encode integers with the same precision as their Rust type, but you don't need to specify it, as it is the default already.

Serde is structural; it doesn't really care about how the data is actually represented in memory. The numerical values you assigned to the enum are lost and bincode will only see the names of the variants.

You can solve this by configuring serde to convert your enum into an integer before exposing it to the bincode serializer:

#[derive(Serialize, Deserialize, Copy, Clone)]
#[serde(into = "u16")]
enum TestEnum {
    Start,
    Init,
    Complete = 65535,
}

impl From<TestEnum> for u16 {
    fn from(value: TestEnum) -> u16 {
        value as u16
    }
}

Then your code gives the expected result:

let x = TestEnum::Complete;
if let Ok(bytes) = bincode::serialize(&x) {
    println!("{:?}", bytes); // [255, 255]
}

thanks so much for the helpful response. I've just tried this and it works exactly as I need. Much appreciated! — Nocker, Oct 24 '20 at 12:09
How would I implement this for an enum with values like `#[repr(u8)] enum Checksum { Crc16([u8; 16]), Crc32([u8; 32]) }` ? — Phidelux, Mar 17 '23 at 09:05

score 0 · Answer 2 · answered Nov 19 '20 at 05:29

I did use the answer provided by @Peter Hall but I had to add an implementation for all of the enums that are defined (there are lots of them) to be able to serialise them and a second implementation to be able to deserialise. So I continued my investigation and found an alternative that seems a little more useable to me. Instead of adding an implementation for each enum, I'm now using the num_enum crate from crates.io and adding some attributes to the enums. So the code looks like below:

use num_enum::{IntoPrimitive, TryFromPrimitive};

#[derive(Serialize, Deserialize, Copy, Clone)]
#[serde(into = "u16", try_from = "u16")]
#[repr(u16)]
enum TestEnum {
    Start,
    Init,
    Complete = 65535,
}

I just thought this might help others in the future.

Rust bincode enum binary serialisation

2 Answers2