谷歌 Go 语言中的并发例程

Question

在go中是否可能:假设，我有3个并发例程，它们可以相互发送整数。现在，假设两个并发例程2 &并发例程3发送一个整数给并发例程1。在go中，例程1是否可能同时取两个值并进一步处理它?为了说清楚，我有以下代码:

package main
import "rand"
func Routine1(command12 chan int, response12 chan int, command13 chan int, response13 chan int ) {
for i := 0; i < 10; i++ {
    i := rand.Intn(100)
if i%2 == 0 {
command12 <- i 
}
if i%2 != 0 {
command13 <- i 
}
print(<-response13, " 1stn");
}
close(command12)
}
func Routine2(command12 chan int, response12 chan int, command23 chan int, response23 chan int) {
for i := 0; ; i++ {
    x, open := <-command12
    if !open {
        return;
    }
     print(x , " 2ndn");
    y := rand.Intn(100)
    if i%2 == 0 {
command12 <- y 
}
if i%2 != 0 {
command23 <- y 
}
}
}
func Routine3(command13 chan int, response13 chan int, command23 chan int, response23 chan int) {
for i := 0; ; i++ {
    x, open := <-command13
    if !open {
        return;
    }
     print(x , " 3ndn");
    y := rand.Intn(100)
    response23 <- y
}
}
func main() {
   command12 := make(chan int)
   response12 := make(chan int)
   command13 := make(chan int)
   response13 := make(chan int)
   command23 := make(chan int)
   response23 := make(chan int)
   go Routine1(command12, response12,command13, response13 )
   Routine2(command12, response12,command23, response23)
   Routine3(command13, response13,command23, response23 )
}

在这个例子中，例程1可以向例程2或3发送一个int值。我想这是例行公事。现在假设，例程3也向例程2发送一个int。例程2是否有可能采用这两个值并进一步处理(动态并发例程)?任何人都可以帮助修改这个程序。

Answer 1

我讨厌抽象的例子，无论如何我会尽力回答你的问题。

是否可能在go中，例程1同时取两个值并进一步处理它?

你想存档什么?在routine1中，你可以这样做:

// Read exactly one command from routine2 as well as exactly
// one command from routine3
cmd1 := <-command12
cmd2 := <-command13
// Process the pair of the two commands here

或

// Process a single command only, which was either sent by routine2
// or by routine3. If there are commands available on both channels
// (command12 and command13) the select statement chooses a branch
// fairly.
select {
    case cmd1 := <-command12:
        // process command from routine 2
    case cmd2 := <-command13
        // process command from routine 3
}

我希望这将回答你的问题。还要注意，Go通道默认情况下支持多个写入器(以及多个读取器)。因此，每个程序只使用一个输入通道可能就足够了。例如，routine1可能只从名为command1的通道读取命令，但routine2和routine3可能都使用相同的command1通道向routine1发送消息。

Go中另一个常见的习惯用法是将回复通道作为消息的一部分传递。例如:

type Command struct {
    Cmd string
    Reply chan-> int
}
func routine2() {
    reply := make(chan int)
    command1 <- Command{"doSomething", reply}
    status := <-reply
}
func routine1() {
    cmd <- command1;
    // process cmd.Cmd
    cmd.Reply <- 200 // SUCCESS (status code)
}

根据您的实际问题，这可能会大大简化您的程序:)

Answer 2

这在GO中是不可能的，我的意思是创建并发通道