Golang 中的并发积分计算

Question

我尝试同时计算积分，但我的程序最终比使用普通 for 循环计算积分要慢。我究竟做错了什么？

package main

import (

    "fmt"

    "math"

    "sync"

    "time"

)

type Result struct {

    result float64

    lock sync.RWMutex

}

var wg sync.WaitGroup

var result Result

func main() {

    now := time.Now()

    a := 0.0

    b := 1.0

    n := 100000.0

    deltax := (b - a) / n

    wg.Add(int(n))

    for i := 0.0; i < n; i++ {

        go f(a, deltax, i)

    }

    wg.Wait()

    fmt.Println(deltax * result.result)

    fmt.Println(time.Now().Sub(now))

}

func f(a float64, deltax float64, i float64) {

    fx := math.Sqrt(a + deltax * (i + 0.5))

    result.lock.Lock()

    result.result += fx

    result.lock.Unlock()

    wg.Done()

}

Answer 1

3-为了提高性能，您可以在不使用的情况下按 CPU 内核划分任务lock sync.RWMutex：+30x使用通道和 进行优化runtime.NumCPU()，这需要2ms2 核和993µs8 核，而您的示例代码需要61ms2 核和40ms8 核：请参阅此工作示例代码和输出：package mainimport (&nbsp; &nbsp; "fmt"&nbsp; &nbsp; "math"&nbsp; &nbsp; "runtime"&nbsp; &nbsp; "time")func main() {&nbsp; &nbsp; nCPU := runtime.NumCPU()&nbsp; &nbsp; fmt.Println("nCPU =", nCPU)&nbsp; &nbsp; ch := make(chan float64, nCPU)&nbsp; &nbsp; startTime := time.Now()&nbsp; &nbsp; a := 0.0&nbsp; &nbsp; b := 1.0&nbsp; &nbsp; n := 100000.0&nbsp;&nbsp; &nbsp; deltax := (b - a) / n&nbsp; &nbsp; stepPerCPU := n / float64(nCPU)&nbsp; &nbsp; for start := 0.0; start < n; {&nbsp; &nbsp; &nbsp; &nbsp; stop := start + stepPerCPU&nbsp; &nbsp; &nbsp; &nbsp; go f(start, stop, a, deltax, ch)&nbsp; &nbsp; &nbsp; &nbsp; start = stop&nbsp; &nbsp; }&nbsp; &nbsp; integral := 0.0&nbsp; &nbsp; for i := 0; i < nCPU; i++ {&nbsp; &nbsp; &nbsp; &nbsp; integral += <-ch&nbsp; &nbsp; }&nbsp; &nbsp; fmt.Println(time.Now().Sub(startTime))&nbsp; &nbsp; fmt.Println(deltax * integral)}func f(start, stop, a, deltax float64, ch chan float64) {&nbsp; &nbsp; result := 0.0&nbsp; &nbsp; for i := start; i < stop; i++ {&nbsp; &nbsp; &nbsp; &nbsp; result += math.Sqrt(a + deltax*(i+0.5))&nbsp; &nbsp; }&nbsp; &nbsp; ch <- result}2核输出：nCPU = 22.0001ms0.66666666859004858核输出：nCPU = 8993µs0.6666666685900456您的示例代码，2 核输出：0.666666668590042461.0035ms您的示例代码，8 核输出：0.666666668590041540.9964ms2-为了获得良好的基准统计数据，请使用大量样本（大 n）：正如您在此处看到的，使用 2 个内核需要2 个内核，但在使用 1 个内核的同110ms一个 CPU 上，这需要： 215msn := 10000000.0使用n := 10000000.0单个 goroutine，请参阅此工作示例代码：package mainimport (&nbsp; &nbsp; "fmt"&nbsp; &nbsp; "math"&nbsp; &nbsp; "time")func main() {&nbsp; &nbsp; now := time.Now()&nbsp; &nbsp; a := 0.0&nbsp; &nbsp; b := 1.0&nbsp; &nbsp; n := 10000000.0&nbsp; &nbsp; deltax := (b - a) / n&nbsp; &nbsp; result := 0.0&nbsp; &nbsp; for i := 0.0; i < n; i++ {&nbsp; &nbsp; &nbsp; &nbsp; result += math.Sqrt(a + deltax*(i+0.5))&nbsp; &nbsp; }&nbsp; &nbsp; fmt.Println(time.Now().Sub(now))&nbsp; &nbsp; fmt.Println(deltax * result)}输出：215.0123ms0.6666666666685884使用n := 10000000.0和 2 个 goroutine，请参阅此工作示例代码：package mainimport (&nbsp; &nbsp; "fmt"&nbsp; &nbsp; "math"&nbsp; &nbsp; "runtime"&nbsp; &nbsp; "time")func main() {&nbsp; &nbsp; nCPU := runtime.NumCPU()&nbsp; &nbsp; fmt.Println("nCPU =", nCPU)&nbsp; &nbsp; ch := make(chan float64, nCPU)&nbsp; &nbsp; startTime := time.Now()&nbsp; &nbsp; a := 0.0&nbsp; &nbsp; b := 1.0&nbsp; &nbsp; n := 10000000.0&nbsp; &nbsp; deltax := (b - a) / n&nbsp; &nbsp; stepPerCPU := n / float64(nCPU)&nbsp; &nbsp; for start := 0.0; start < n; {&nbsp; &nbsp; &nbsp; &nbsp; stop := start + stepPerCPU&nbsp; &nbsp; &nbsp; &nbsp; go f(start, stop, a, deltax, ch)&nbsp; &nbsp; &nbsp; &nbsp; start = stop&nbsp; &nbsp; }&nbsp; &nbsp; integral := 0.0&nbsp; &nbsp; for i := 0; i < nCPU; i++ {&nbsp; &nbsp; &nbsp; &nbsp; integral += <-ch&nbsp; &nbsp; }&nbsp; &nbsp; fmt.Println(time.Now().Sub(startTime))&nbsp; &nbsp; fmt.Println(deltax * integral)}func f(start, stop, a, deltax float64, ch chan float64) {&nbsp; &nbsp; result := 0.0&nbsp; &nbsp; for i := start; i < stop; i++ {&nbsp; &nbsp; &nbsp; &nbsp; result += math.Sqrt(a + deltax*(i+0.5))&nbsp; &nbsp; }&nbsp; &nbsp; ch <- result}输出：nCPU = 2110.0063ms0.66666666666860731- Goroutines 的数量有一个最佳点，从这一点开始，增加 Goroutines 的数量并不会减少程序执行时间：在 2 核 CPU 上，使用以下代码，结果是：nCPU: 1,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 4,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;8,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;16Time: 2.1601236s, 1.1220642s, 1.1060633s, 1.1140637s, 1.1380651s正如你所看到的nCPU=1 ，nCPU=2减少量已经足够大，但在此之后它并不多，所以nCPU=22 Cores CPU 是此示例代码的最佳点，所以在这里使用 nCPU := runtime.NumCPU()就足够了。package mainimport (&nbsp; &nbsp; "fmt"&nbsp; &nbsp; "math"&nbsp; &nbsp; "time")func main() {&nbsp; &nbsp; nCPU := 2 //2.1601236s@1 1.1220642s@2 1.1060633s@4 1.1140637s@8 1.1380651s@16&nbsp; &nbsp; fmt.Println("nCPU =", nCPU)&nbsp; &nbsp; ch := make(chan float64, nCPU)&nbsp; &nbsp; startTime := time.Now()&nbsp; &nbsp; a := 0.0&nbsp; &nbsp; b := 1.0&nbsp; &nbsp; n := 100000000.0&nbsp; &nbsp; deltax := (b - a) / n&nbsp; &nbsp; stepPerCPU := n / float64(nCPU)&nbsp; &nbsp; for start := 0.0; start < n; {&nbsp; &nbsp; &nbsp; &nbsp; stop := start + stepPerCPU&nbsp; &nbsp; &nbsp; &nbsp; go f(start, stop, a, deltax, ch)&nbsp; &nbsp; &nbsp; &nbsp; start = stop&nbsp; &nbsp; }&nbsp; &nbsp; integral := 0.0&nbsp; &nbsp; for i := 0; i < nCPU; i++ {&nbsp; &nbsp; &nbsp; &nbsp; integral += <-ch&nbsp; &nbsp; }&nbsp; &nbsp; fmt.Println(time.Now().Sub(startTime))&nbsp; &nbsp; fmt.Println(deltax * integral)}func f(start, stop, a, deltax float64, ch chan float64) {&nbsp; &nbsp; result := 0.0&nbsp; &nbsp; for i := start; i < stop; i++ {&nbsp; &nbsp; &nbsp; &nbsp; result += math.Sqrt(a + deltax*(i+0.5))&nbsp; &nbsp; }&nbsp; &nbsp; ch <- result}

Answer 2

除非 goroutine 中的活动花费的时间比切换上下文、执行任务和使用互斥锁更新值所需的时间多得多，否则串行执行会更快。看一个稍微修改过的版本。我所做的只是在f()函数中添加 1 微秒的延迟。package mainimport (&nbsp; &nbsp; "fmt"&nbsp; &nbsp; "math"&nbsp; &nbsp; "sync"&nbsp; &nbsp; "time")type Result struct {&nbsp; &nbsp; result float64&nbsp; &nbsp; lock&nbsp; &nbsp;sync.RWMutex}var wg sync.WaitGroupvar result Resultfunc main() {&nbsp; &nbsp; fmt.Println("concurrent")&nbsp; &nbsp; concurrent()&nbsp; &nbsp; result.result = 0&nbsp; &nbsp; fmt.Println("serial")&nbsp; &nbsp; serial()}func concurrent() {&nbsp; &nbsp; now := time.Now()&nbsp; &nbsp; a := 0.0&nbsp; &nbsp; b := 1.0&nbsp; &nbsp; n := 100000.0&nbsp; &nbsp; deltax := (b - a) / n&nbsp; &nbsp; wg.Add(int(n))&nbsp; &nbsp; for i := 0.0; i < n; i++ {&nbsp; &nbsp; &nbsp; &nbsp; go f(a, deltax, i, true)&nbsp; &nbsp; }&nbsp; &nbsp; wg.Wait()&nbsp; &nbsp; fmt.Println(deltax * result.result)&nbsp; &nbsp; fmt.Println(time.Now().Sub(now))}func serial() {&nbsp; &nbsp; now := time.Now()&nbsp; &nbsp; a := 0.0&nbsp; &nbsp; b := 1.0&nbsp; &nbsp; n := 100000.0&nbsp; &nbsp; deltax := (b - a) / n&nbsp; &nbsp; for i := 0.0; i < n; i++ {&nbsp; &nbsp; &nbsp; &nbsp; f(a, deltax, i, false)&nbsp; &nbsp; }&nbsp; &nbsp; fmt.Println(deltax * result.result)&nbsp; &nbsp; fmt.Println(time.Now().Sub(now))}func f(a, deltax, i float64, concurrent bool) {&nbsp; &nbsp; time.Sleep(1 * time.Microsecond)&nbsp; &nbsp; fx := math.Sqrt(a + deltax*(i+0.5))&nbsp; &nbsp; if concurrent {&nbsp; &nbsp; &nbsp; &nbsp; result.lock.Lock()&nbsp; &nbsp; &nbsp; &nbsp; result.result += fx&nbsp; &nbsp; &nbsp; &nbsp; result.lock.Unlock()&nbsp; &nbsp; &nbsp; &nbsp; wg.Done()&nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; result.result += fx&nbsp; &nbsp; }}加上延迟，结果如下（并发版本快很多）：concurrent0.6666666685900424624.914165msserial0.66666666859004225.609195767s事不宜迟：concurrent0.666666668590042850.771275msserial0.6666666685900422749.166µs正如你所看到的，完成一项任务所花费的时间越长，如果可能的话，同时执行它就越有意义。