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Catching panic! when Rust called from C FFI, without spawning threads

I'm working on a Rust wrapper for the Duktape JavaScript interpreter. In a normal use case, the call stack will look like this:

  1. Rust: Arbitrary application code.
  2. Rust: My library wrapper.
  3. C: The Duktape interpreter.
  4. Rust: My Rust code.
  5. Rust: Arbitrary callbacks into application code.

What happens if (5) calls panic!? According to various Rust developers on IRC, attempting to panic! from inside non-Rust callframes like (3) may result in undefined behavior.

But according the Rust documentation, the only way to catch a panic! is using std::task::try, which spawns an extra thread. There's also rustrt::unwind::try, which cannot be nested twice within a single thread, among other restrictions.

One solution, proposed by Benjamin Herr, is to abort the process if the code in (5) panics. I've packaged his solution as abort_on_panic, and it appears to work, for values of "work" that include "crashing the entire program, but at least not corrupting things subtly":

abort_on_panic!("cannot panic inside this block", {
    panic!("something went wrong!");
});

But is a way to emulate std::task::try without the overhead of thread/task creation?

like image 924
emk Avatar asked Dec 09 '14 17:12

emk


2 Answers

As of Rust 1.9.0, you can use panic::catch_unwind to recover the error:

use std::panic;

fn main() {
    let result = panic::catch_unwind(|| {
        panic!("oh no!");
    });
    assert!(result.is_err());
}

Passing it to the next layer is just as easy with panic::resume_unwind:

use std::panic;

fn main() {
    let result = panic::catch_unwind(|| {
        panic!("oh no!");
    });

    if let Err(e) = result {
        panic::resume_unwind(e);
    }
}
like image 181
dragostis Avatar answered Oct 27 '22 00:10

dragostis


Editor's note: This answer predates Rust 1.0 and is no longer necessarily accurate. Other answers still contain valuable information.

You cannot 'catch' a panic!. It terminates execution of the current thread. Therefore, without spinning up a new one to isolate, it's going to terminate the thread you're in.

like image 27
Steve Klabnik Avatar answered Oct 27 '22 01:10

Steve Klabnik