send feature not working

[ezenkico]

Hi guys

I'm having trouble getting the send safe setup working.

I have the following code.

async fn run_base(
    input: Vec<u8>
) -> mlua::prelude::LuaResult<()>{
    let lua = Lua::new_with(StdLib::TABLE | StdLib::STRING | StdLib::MATH, LuaOptions::new())?;

    let map_table = lua.create_table()?;
    map_table.set(1, "one")?;
    map_table.set("two", 2)?;

    lua.globals().set("map_table", map_table)?;

    lua.load(input).exec_async().await?;

    Ok(())
}

fn run_lua(
    input: Vec<u8>
) -> Pin<Box<impl Future<Output = mlua::prelude::LuaResult<()>> + Send>>{
    Box::pin(async move{
        run_base(input)
    })
}

and this is what I have in cargo

mlua = {version="0.9.5", features = ["async", "send", "vendored", "luau"]}

And I'm getting the following errors

141 | ) -> Pin<Box<impl Future<Output = mlua::prelude::LuaResult<()>> + Send>>{ | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^*mut lua_Statecannot be sent between threads safely 141 | ) -> Pin<Box<impl Future<Output = mlua::prelude::LuaResult<()>> + Send>>{ | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^UnsafeCellmlua::lua::ExtraDatacannot be shared between threads safely 141 | ) -> Pin<Box<impl Future<Output = mlua::prelude::LuaResult<()>> + Send>>{ | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^*const c_voidcannot be sent between threads safely 141 | ) -> Pin<Box<impl Future<Output = mlua::prelude::LuaResult<()>> + Send>>{ | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^NonNullcannot be sent between threads safely | 141 | ) -> Pin<Box<impl Future<Output = mlua::prelude::LuaResult<()>> + Send>>{ | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^(dyn for<'a> Fn(&'a Lua) -> Result<LuaVmState, LuaError> + Send + 'static)cannot be shared between threads safely 141 | ) -> Pin<Box<impl Future<Output = mlua::prelude::LuaResult<()>> + Send>>{ | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^*mut c_voidcannot be sent between threads safely

The only other possible option I've seen was to use threads and that can get expensive.

If someone can help me out I would appreciate it.

[khvzak]

send feature works but probably not the way you expect.
You're getting this error becase future returned from exec_async is !Send. Making it sendable would require making Lua types Sync which is not yet implemented.

Maybe if you tell what you want to achieve I can suggest an alternative solution. Eg. in tokio you can spawn non-Send futures usuing LocalSet

[ezenkico]

I appreciate the response.

I can't go into to much detail but in a nutshell I am trying to run lua code on demand and due to other parts of the code I need to do this in an async function that is send safe. The only thing I can think of is passing things into another thread which will be costly and I would prefer to avoid that.

If you could help that would be great.

Thanks

[khvzak]

Even to use LocalSet you would need to run another thread. In general passing data to threads is cheap, as all threads share the same address space. If you can spawn a dedicated thread for Lua and then make communication through mpsc channel, it should be sufficient. You can keep references to Lua values using RegistryKey, they are Send+Sync.

[khvzak]

An example of using channels

#[tokio::main]
async fn main() -> mlua::Result<()> {
    let chunk_tx = {
        let (send, mut recv) =
            tokio::sync::mpsc::unbounded_channel::<(String, Sender<LuaResult<()>>)>();
        let rt = Builder::new_current_thread().enable_all().build().unwrap();
        std::thread::spawn(move || {
            rt.block_on(async {
                let lua = Lua::new();
                while let Some((chunk, res_tx)) = recv.recv().await {
                    let res = lua.load(chunk).exec_async().await;
                    res_tx.send(res).unwrap();
                }
            });
        });
        send
    };

    let (res_tx, res_rx) = tokio::sync::oneshot::channel();
    chunk_tx
        .send(("print('Hello, World!')".into(), res_tx))
        .unwrap();
    let res = res_rx.await.unwrap()?;
    println!("Result: {:?}", res);

    Ok(())
}

[ezenkico]

Yeah, I had a similar idea. I was just hoping I could avoid it.
If it's not too much trouble, in a future iteration, if you could make it sendable that would be great. I'll work with this for now though.

I appreciate your help on this.

Thanks.