TCP到Redis服务器且通道卡死

Question

I need to send echo aaa three times to redis server, but its get stuck in the middle of process, I also check if read and write operation get error message, but it doesn't. so, why it get stuck in the middle of process?

package main



import (
    "fmt"
    "os"
    "io"
    "net"
    "sync"
)

var (
    wg = new(sync.WaitGroup)
)

func readFromServer(isWrite chan bool, r io.Reader) {
    for {
        select {
        case <-isWrite:
            _ , err := io.Copy(os.Stdout, r)
            if err != nil {
                panic(err)
            }
        }
    }
}

func writeToServer(conn net.Conn , isWrite chan bool ){
    defer wg.Done()
    for i :=0; i<3; i++{
        _ , err := conn.Write([]byte("*2
$4
echo
$3
aaa
"))
        if err != nil {
            panic(err)
        }
        isWrite<- true
    }
}

func main(){
    wg.Add(1)

    conn ,err := net.Dial("tcp","127.0.0.1:6379")
    isWrite := make(chan bool)

    if err != nil {
        panic(err)
    }

    go readFromServer(isWrite, conn)
    go writeToServer(conn , isWrite)



    wg.Wait()
    fmt.Println("finished...")
}

Output:

$3
aaa
$3
aaa
Stuck here...

Answer 1

The readFromServer function receives one value from the isWrite channel and then blocks in the call to io.Copy. The io.Copy function does not return until EOF or some error reading or writing data. All of the program output is from the single call to io.Copy.

The second send to isWrite in sendToServer blocks. The isWrite channel is an unbuffered channel. A send on an unbuffered channel does not proceed until there's a receiver. There is no receiver on the channel because readFromServer is blocked in the call to io.Copy.

Possible fixes are:

• The fix is to modify readFromServer to parse the RESP protocol and read exactly one message per iteration in the loop.

• Replace for loop in readFromServer with a single call to io.Copy.

The isWrite channel is not needed.

The program does not ensure that readFromServer reads all of the responses from writeToServer before the program exits.