c - Rust 与 C 性能

Question

我想学习一些关于 Rust 任务的知识，所以我做了一个蒙特卡洛计算圆周率。现在我的困惑是为什么单线程 C 版本快 4 倍比 4 路线程 Rust 版本。很明显我做错了什么，或者我的心智表现模式有偏差。

这是 C 版本:

#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>
#include <stdio.h>

#define PI 3.1415926535897932

double monte_carlo_pi(int nparts)
{
    int i, in=0;
    double x, y;
    srand(getpid());

    for (i=0; i<nparts; i++) {
        x = (double)rand()/(double)RAND_MAX;
        y = (double)rand()/(double)RAND_MAX;

            if (x*x + y*y < 1.0) {
            in++;
        }
    }

    return in/(double)nparts * 4.0;
}

int main(int argc, char **argv)
{
    int nparts;
    double mc_pi;

    nparts = atoi(argv[1]);
    mc_pi = monte_carlo_pi(nparts);
    printf("computed: %f error: %f\n", mc_pi, mc_pi - PI);
}

Rust 版本不是逐行移植:

use std::rand;
use std::rand::distributions::{IndependentSample,Range};

fn monte_carlo_pi(nparts: uint ) -> uint {
    let between = Range::new(0f64,1f64);
    let mut rng = rand::task_rng();
    let mut in_circle = 0u;
    for _ in range(0u, nparts) {
        let a = between.ind_sample(&mut rng);
    let b = between.ind_sample(&mut rng);

    if a*a + b*b <= 1.0 {
        in_circle += 1;
    }
    }
    in_circle
}

fn main() {
    let (tx, rx) = channel();

    let ntasks = 4u;
    let nparts = 100000000u; /* I haven't learned how to parse cmnd line args yet!*/
    for _ in range(0u, ntasks) {
        let child_tx = tx.clone();
        spawn(proc() {
        child_tx.send(monte_carlo_pi(nparts/ntasks));
        });
    }

    let result = rx.recv() + rx.recv() + rx.recv() + rx.recv();

    println!("pi is {}", (result as f64)/(nparts as f64)*4.0);
}

构建并计时 C 版本:

$ clang -O2 mc-pi.c -o mc-pi-c; time ./mc-pi-c 100000000
computed: 3.141700 error: 0.000108
./mc-pi-c 100000000  1.68s user 0.00s system 99% cpu 1.683 total

构建并计时 Rust 版本:

$ rustc -v      
rustc 0.12.0-nightly (740905042 2014-09-29 23:52:21 +0000)
$ rustc --opt-level 2 --debuginfo 0 mc-pi.rs -o mc-pi-rust; time ./mc-pi-rust  
pi is 3.141327
./mc-pi-rust  2.40s user 24.56s system 352% cpu 7.654 tota

Answer 1

正如 Dogbert 观察到的那样，瓶颈在于随机数生成器。这是一个快速的并且在每个线程上播种不同的

fn monte_carlo_pi(id: u32, nparts: uint ) -> uint {
    ...
    let mut rng: XorShiftRng = SeedableRng::from_seed([id,id,id,id]);
    ...
}