go - 发送到 channel 时避免竞争条件？

Question

go版本go1.11.2 darwin/amd64

我有以下代码示例，是为 SO 演示目的而创建的:

package main

import (
    ...
)

type T struct {
    ctx context.Context
    ch1 chan string
}

func New(ctx context.Context) *T {
    t := &T{ctx: ctx}
    go t.run(2)
    return t

}

func (t *T) run(workers int) {
    t.ch1 = make(chan string)
    done := make(chan struct{})

    go func() {
        <-t.ctx.Done()
        close(done)
        close(t.ch1)
    }()

    for i := 0; i < workers; i++ {
        go func() {
            for {
                select {
                case <-done:
                    return
                case m, ok := <-t.ch1:
                    if ok {
                        t.process(done, m)
                    }
                }
            }
        }()
    }
}

func (t *T) process(done <-chan struct{}, s string) {
    select {
    case <-done:
        return
    default:
        log.Printf("processing %s", s)
        time.Sleep(time.Millisecond * 200)
    }
}

func (t *T) Read() <-chan string {
    return t.ch1
}

func (t *T) Write(s string) error {
    select {
    case <-t.ctx.Done():
        return errors.New("consumer is closed today")
    case t.ch1 <- s:
        return nil
    }
}

func main() {
    ctx, cancel := context.WithCancel(context.Background())
    t := New(ctx)

    go func() {
        for m := range t.Read() {
            log.Printf("got %s", m)
        }
        <-ctx.Done()
    }()

    for i := 0; i < 10; i++ {
        t.Write(strconv.Itoa(i))
    }
    cancel()
}

当我使用竞争检测器构建并运行它时，它会抛出以下数据竞争:

go build -race ./test/ && ./test
==================
WARNING: DATA RACE
Read at 0x00c0000b6030 by goroutine 7:
  main.main.func1()
      /redacted/test/app.go:60 +0x42

Previous write at 0x00c0000b6030 by goroutine 6:
  main.(*T).run()
      /redacted/test/app.go:24 +0x6a

Goroutine 7 (running) created at:
  main.main()
      /redacted/test/app.go:76 +0xbc

Goroutine 6 (running) created at:
  main.New()
      /redacted/test/app.go:18 +0xcd
  main.main()
      /redacted/test/app.go:74 +0x86
==================
==================
WARNING: DATA RACE
Read at 0x00c0000b6030 by main goroutine:
  main.(*T).Write()
      /redacted/test/app.go:67 +0x8a
  main.main()
      /redacted/test/app.go:84 +0xdc

Previous write at 0x00c0000b6030 by goroutine 6:
  main.(*T).run()
      /redacted/test/app.go:24 +0x6a

Goroutine 6 (running) created at:
  main.New()
      /redacted/test/app.go:18 +0xcd
  main.main()
      /redacted/test/app.go:74 +0x86
==================
2019/01/20 10:48:51 got 0
2019/01/20 10:48:51 got 3
2019/01/20 10:48:51 processing 1
2019/01/20 10:48:51 processing 2
2019/01/20 10:48:51 got 4
2019/01/20 10:48:51 got 5
2019/01/20 10:48:51 got 6
2019/01/20 10:48:51 got 7
2019/01/20 10:48:51 got 8
2019/01/20 10:48:51 got 9
Found 2 data race(s)

我遇到的问题是，我似乎无法找到一种方法让用户在 channel 中输入内容，而不暴露任何写入 channel ，没有竞争。这怎么可能？有没有我缺少的更好的模式？

Answer 1

我建议进行以下更改:

• 分配给 New 中的 ch1 以避免在多个 goroutine 中读取和写入 t.ch1 的竞争
• 仅在所有对Write 的调用完成后关闭ch1，以避免“在已关闭的 channel 上发送” panic
• 使用 sync.WaitGroup 在写入所有值后等待所有处理 goroutine 完成(这样程序不会在处理完成之前退出)

将这些变化结合在一起，它看起来是这样的:

package main

import (
    "log"
    "strconv"
    "sync"
    "time"
)

type T struct {
    // ch1 receives the values to process
    ch1 chan string

    // wg is used to wait for the workers to stop
    wg sync.WaitGroup
}

func New() *T {
    t := &T{
        ch1: make(chan string),
    }
    go t.run(2)
    return t
}

func (t *T) run(workers int) {
    // add the workers to the WaitGroup
    t.wg.Add(workers)

    for i := 0; i < workers; i++ {
        go func() {
            // process values from the channel until it closes
            // and then signal to the WaitGroup that we're done
            defer t.wg.Done()
            for m := range t.ch1 {
                t.process(m)
            }
        }()
    }
}

// Stop is called after we're done calling Write and we want to stop the
// processing once all values have been processed
func (t *T) Stop() {
    // close t.ch1 so that the workers know to stop processing
    close(t.ch1)

    // wait for the workers to all finish before returning
    t.wg.Wait()
}

func (t *T) process(s string) {
    log.Printf("processing %s", s)
    time.Sleep(time.Millisecond * 200)
}

func (t *T) Write(s string) {
    t.ch1 <- s
}

func main() {
    // start the main loop
    t := New()

    // write 10 values
    for i := 0; i < 10; i++ {
        t.Write(strconv.Itoa(i))
    }

    // stop the loop, which will wait for processing to finish before returning
    t.Stop()
}