Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
710 views
in Technique[技术] by (71.8m points)

rust - Simultaneous mutable access to arbitrary indices of a large vector that are guaranteed to be disjoint

Context

I have a case where multiple threads must update objects stored in a shared vector. However, the vector is very large, and the number of elements to update is relatively small.

Problem

In a minimal example, the set of elements to update can be identified by a (hash-)set containing the indices of elements to update. The code could hence look as follows:

let mut big_vector_of_elements = generate_data_vector();

while has_things_to_do() {
    let indices_to_update = compute_indices();
    indices_to_update.par_iter() // Rayon parallel iteration
       .map(|index| big_vector_of_elements[index].mutate())
       .collect()?;
}

This is obviously disallowed in Rust: big_vector_of_elements cannot be borrowed mutably in multiple threads at the same time. However, wrapping each element in, e.g., a Mutex lock seems unnecessary: this specific case would be safe without explicit synchronization. Since the indices come from a set, they are guaranteed to be distinct. No two iterations in the par_iter touch the same element of the vector.

Restating my question

What would be the best way of writing a program that mutates elements in a vector in parallel, where the synchronization is already taken care of by the selection of indices, but where the compiler does not understand the latter?

A near-optimal solution would be to wrap all elements in big_vector_of_elements in some hypothetical UncontendedMutex lock, which would be a variant of Mutex which is ridiculously fast in the uncontended case, and which may take arbitrarily long when contention occurs (or even panics). Ideally, an UncontendedMutex<T> should also be of the same size and alignment as T, for any T.

Related, but different questions:

Multiple questions can be answered with "use Rayon's parallel iterator", "use chunks_mut", or "use split_at_mut":

These answers do not seem relevant here, since those solutions imply iterating over the entire big_vector_of_elements, and then for each element figuring out whether anything needs to be changed. Essentially, this means that such a solution would look as follows:

let mut big_vector_of_elements = generate_data_vector();

while has_things_to_do() {
    let indices_to_update = compute_indices();
    for (index, mut element) in big_vector_of_elements.par_iter().enumerate() {
        if indices_to_update.contains(index) {
            element.mutate()?;
        }
    }
}

This solution takes time proportionate to the size of big_vector_of_elements, whereas the first solution loops only over a number of elements proportionate to the size of indices_to_update.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

You can sort indices_to_update and extract mutable references by calling split_*_mut.

let len = big_vector_of_elements.len();

while has_things_to_do() {
    let mut tail = big_vector_of_elements.as_mut_slice();

    let mut indices_to_update = compute_indices();
    // I assumed compute_indices() returns unsorted vector
    // to highlight the importance of sorted order
    indices_to_update.sort();

    let mut elems = Vec::new();

    for idx in indices_to_update {
        // cut prefix, so big_vector[idx] will be tail[0]
        tail = tail.split_at_mut(idx - (len - tail.len())).1;

        // extract tail[0]
        let (elem, new_tail) = tail.split_first_mut().unwrap();
        elems.push(elem);

        tail = new_tail;
    }
}

Double check everything in this code; I didn't test it. Then you can call elems.par_iter(...) or whatever.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...