Do tasks inherit tracing spans?

[BGR360]

When a task is spawned, does it inherit the current tracing span? I don't see this called out in the docs, nor am I able to easily figure it out by reading the tokio source code. I (think that I) want this behavior, because I want to be able to associate spawned tasks with the original operations with which they were meant to help.

[hawkw]

Tasks don't inherit the current tracing span unless the current span is explicitly propagated to the spawned task. For example, if you want to spawn a task in the same span as the context in which it's spawned, you can use Instrument::in_current_span:

use tracing::Instrument;

tokio::spawn(my_task().in_current_span());
// which is equivalent to
tokio::spawn(my_task().instrument(tracing::Span::current());

Alternatively, if you instrument the spawned future with a new span, that span will have the current span as its parent, so that task will be conceptually "inside" the current span:

use tracing::{Instrument, info_span};

// creates a new span called `my_task`, which is a child of the current span, and
// attaches it to the spawned future.
tokio::spawn(my_task().instrument(info_span!("my_task"));

Finally, if you want a spawned task to inherit the current task's span if the child span is not enabled by filtering, you can use Span::or_current:

use tracing::{Instrument, info_span};

// creates a new span called `my_task`, which is a child of the current span:
let span = info_span!("my_task")
     // if the child span is not enabled by the `tracing` subscriber, use the 
     // current span, instead:
     .or_current();
tokio::spawn(my_task().instrument(span));

[BGR360]

Heck yes, that is so helpful and cool. Thank you!

[BGR360]

If the future I'm spawning is the output of an #[instrument]-ed async fn, will it have the current span as its parent without me calling .instrument() on it? I thought I remembered seeing something about #[instrument] spitting out Instrumented futures already.

[hawkw]

Yes, the span generated by #[instrument] will have the current span when the async fn is called as its parent. Putting #[instrument] on an async fn is essentially saying "every time I call this async fn, I want to immediately .instrument() it with this span".