• 4 Posts
  • 48 Comments
Joined 2 years ago
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Cake day: August 5th, 2023

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  • A closure may/will try to capture by reference, so it may hold references to the calling function’s scope. For example, this would want to hold a reference to x:

    let x = 0;
    std::thread::spawn(|| x += 1);
    

    It’s as if you had a struct like this:

    struct MyClosure<'a> {
        x: &'a mut i32,
    }
    

    That makes the closure non-'static, since it holds non-'static references. The usual solution is to use the move keyword, to hint that you want it to move by default, or to use scoped threads. But yeah Send and 'static depend on the captures.

    Am I correct in guessing that you handle is of Gontian origin?

    Yes! 😁 I picked it when I used to play Tibia (15-20 years ago!), and it stuck with me since then. The correct spelling was already taken, so I modified it a bit. This name is usually available.


  • Your guess is correct, it should be understood as

    F: ( FnOnce() -> T ) + Send + 'static
    T: Send + 'static
    

    The FnOnce() -> T is syntax sugar for FnOnce<(), Output = T> and the bounds after apply to F. That’s also why T has separate bounds. They aren’t propagated or inherited. It’s just an ambiguous looking syntax, but T + Trait + 'lifetime isn’t a valid syntax for applying bounds to T (unless I missed something).

    The type F may be a closure over values from the calling thread. It has to be possible to send these values to the spawned thread, hence F needs Send + 'static. When the thread is done with its work, it returns a value of type T that gets passed back via JoinHandle<T>, so T needs Send + 'static too.

    I hope this cleared things up a bit.