我正在并行填充向量,但对于这个广义问题,我只发现了提示而没有答案。
下面的代码有效,但我想切换到Rng::fill
而不是迭代每个块。在单个Vec
中可能不可能有多个可变切片;我不确定。
extern crate rayon;
extern crate rand;
extern crate rand_xoshiro;
use rand::{Rng, SeedableRng};
use rand_xoshiro::Xoshiro256StarStar;
use rayon::prelude::*;
use std::{iter, env};
use std::sync::{Arc, Mutex};
// i16 because I was filling up my RAM for large tests...
fn gen_rand_vec(data: &mut [i16]) {
let num_threads = rayon::current_num_threads();
let mut rng = rand::thread_rng();
let mut prng = Xoshiro256StarStar::from_rng(&mut rng).unwrap();
// lazy iterator of fast, unique RNGs
// Arc and Mutex are just so it can be accessed from multiple threads
let rng_it = Arc::new(Mutex::new(iter::repeat(()).map(|()| {
let new_prng = prng.clone();
prng.jump();
new_prng
})));
// generates random numbers for each chunk in parallel
// par_chunks_mut is parallel version of chunks_mut
data.par_chunks_mut(data.len() / num_threads).for_each(|chunk| {
// I used extra braces because it might be required to unlock Mutex.
// Not sure.
let mut prng = { rng_it.lock().unwrap().next().unwrap() };
for i in chunk.iter_mut() {
*i = prng.gen_range(-1024, 1024);
}
});
}
事实证明,ChunksMut
迭代器给出了切片。我不确定如何从文档中收集到。我通过阅读the source想出来:
#[derive(Debug)]
#[stable(feature = "rust1", since = "1.0.0")]
pub struct ChunksMut<'a, T:'a> {
v: &'a mut [T],
chunk_size: usize
}
#[stable(feature = "rust1", since = "1.0.0")]
impl<'a, T> Iterator for ChunksMut<'a, T> {
type Item = &'a mut [T];
#[inline]
fn next(&mut self) -> Option<&'a mut [T]> {
if self.v.is_empty() {
None
} else {
let sz = cmp::min(self.v.len(), self.chunk_size);
let tmp = mem::replace(&mut self.v, &mut []);
let (head, tail) = tmp.split_at_mut(sz);
self.v = tail;
Some(head)
}
}
我想这只是经验;对于其他人来说,显而易见的是,来自ChunksMut<T>
的Vec<T>
类型的迭代器返回[T]
类型的对象。这是有道理的。中间结构只是不太清楚。
pub fn chunks_mut(&mut self, chunk_size: usize) -> ChunksMut<T>
// ...
impl<'a, T> Iterator for ChunksMut<'a, T>
读这个,它看起来像迭代器返回类型为T
的对象,与Vec<T>.iter()
相同,这是没有意义的。