Go: One producer many consumers

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情话喂你
情话喂你 2021-02-02 13:56

So I have seen a lot of ways of implementing one consumer and many producers in Go - the classic fanIn function from the Concurrency in Go talk.

What I want is a fanOut

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4条回答
  • 2021-02-02 14:03

    We can handle multiple consumers without making the copy of channel data for each consumer.

    Go playground: https://play.golang.org/p/yOKindnqiZv

    package main
    
    import (
        "fmt"
        "sync"
    )
    
    type data struct {
        msg string
        consumers int
    }
    
    func main() {
        ch := make(chan *data) // both block or non-block are ok
        var wg sync.WaitGroup
        consumerCount := 3 // specify no. of consumers
    
        producer := func() {
            obj := &data {
                msg: "hello everyone!",
                consumers: consumerCount,
            }
            ch <- obj
        }
        consumer := func(idx int) {
            defer wg.Done()
            obj := <-ch
            fmt.Printf("consumer %d received data %v\n", idx, obj)
            obj.consumers--
            if obj.consumers > 0 {
                ch <- obj // forward to others
            } else {
                fmt.Printf("last receiver: %d\n", idx)
            }
        }
    
        go producer()
        for i:=1; i<=consumerCount; i++ {
            wg.Add(1)
            go consumer(i)
        }
    
        wg.Wait()
    }
    
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  • 2021-02-02 14:06

    You pretty much described the best way to do it but here is a small sample of code that does it.

    Go playground: https://play.golang.org/p/jwdtDXVHJk

    package main
    
    import (
        "fmt"
        "time"
    )
    
    func producer(iters int) <-chan int {
        c := make(chan int)
        go func() {
            for i := 0; i < iters; i++ {
                c <- i
                time.Sleep(1 * time.Second)
            }
            close(c)
        }()
        return c
    }
    
    func consumer(cin <-chan int) {
        for i := range cin {
            fmt.Println(i)
        }
    }
    
    func fanOut(ch <-chan int, size, lag int) []chan int {
        cs := make([]chan int, size)
        for i, _ := range cs {
            // The size of the channels buffer controls how far behind the recievers
            // of the fanOut channels can lag the other channels.
            cs[i] = make(chan int, lag)
        }
        go func() {
            for i := range ch {
                for _, c := range cs {
                    c <- i
                }
            }
            for _, c := range cs {
                // close all our fanOut channels when the input channel is exhausted.
                close(c)
            }
        }()
        return cs
    }
    
    func fanOutUnbuffered(ch <-chan int, size int) []chan int {
        cs := make([]chan int, size)
        for i, _ := range cs {
            // The size of the channels buffer controls how far behind the recievers
            // of the fanOut channels can lag the other channels.
            cs[i] = make(chan int)
        }
        go func() {
            for i := range ch {
                for _, c := range cs {
                    c <- i
                }
            }
            for _, c := range cs {
                // close all our fanOut channels when the input channel is exhausted.
                close(c)
            }
        }()
        return cs
    }
    
    func main() {
        c := producer(10)
        chans := fanOutUnbuffered(c, 3)
        go consumer(chans[0])
        go consumer(chans[1])
        consumer(chans[2])
    }
    

    The important part to note is how we close the output channels once the input channel has been exhausted. Also if one of the output channels blocks on the send it will hold up the send on the other output channels. We control the amount of lag by setting the buffer size of the channels.

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  • 2021-02-02 14:07

    First, see related question What is the neatest idiom for producer/consumer in Go? and One thread showing interest in another thread (consumer / producer). Also, take a look to producer-consumer problem. About concurrency see how to achieve concurrency In Google Go.

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  • 2021-02-02 14:15

    This solution below is a bit contrived, but it works for me:

    package main
    
    import (
        "fmt"
        "time"
        "crypto/rand"
        "encoding/binary"
    )
    
    func handleNewChannels(arrchangen chan [](chan uint32),
                           intchangen chan (chan uint32)) {
        currarr := []chan uint32{}
        arrchangen <- currarr
        for {
            newchan := <-intchangen
            currarr = append(currarr, newchan)
            arrchangen <- currarr
        }
    }
    
    func sendToChannels(arrchangen chan [](chan uint32)) {
        tick := time.Tick(1 * time.Second)
        currarr := <-arrchangen
        for {
            select {
            case <-tick:
                sent := false
                var n uint32
                binary.Read(rand.Reader, binary.LittleEndian, &n)
                for i := 0 ; i < len(currarr) ; i++ {
                    currarr[i] <- n
                    sent = true
                }
                if sent {
                    fmt.Println("Sent generated ", n)
                }
            case newarr := <-arrchangen:
                currarr = newarr
            }
        }
    }
    func handleChannel(tchan chan uint32) {
        for {
            val := <-tchan
            fmt.Println("Got the value ", val)
        }
    }
    
    func createChannels(intchangen chan (chan uint32)) {
        othertick := time.Tick(5 * time.Second)
        for {
            <-othertick
            fmt.Println("Creating new channel! ")
            newchan := make(chan uint32)
            intchangen <- newchan
            go handleChannel(newchan)
        }
    }
    
    func main() {
        arrchangen := make(chan [](chan uint32))
        intchangen := make(chan (chan uint32))
        go handleNewChannels(arrchangen, intchangen)
        go sendToChannels(arrchangen)
        createChannels(intchangen)
    }
    
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