Question

是否可以去：假设，我有3个并发例程，可以相互发送整数。现在，假设两个并发的例程2＆amp; 3发送一个整数到并发例程1.是否有可能在例程1中取两个值并进一步处理它？为清楚起见，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, " 1st\n");
}
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 , " 2nd\n");
    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 , " 3nd\n");
    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 )
}

在这个示例中，例程1可以向例程2或3发送一个int。我假设它是例程3.现在假设，例程3也向例程2发送一个int。例程2是否可以采用这两个例程。值和进一步处理（动态并发例程）？任何机构都可以帮助相应地修改此程序。

Answer 1

我讨厌抽象的例子，无论如何我会尽力回答你的问题。

是否有可能在例程1中同时获取两个值并将其处理得更远？

您想要存档什么？在例程1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

OR

// 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
}

我希望能回答你的问题。另请注意，默认情况下，Go通道支持多个写入器（以及多个读取器）。因此，每个goroutine只使用一个输入通道就足够了。例如，routine1可能只读取来自名为command1的通道的命令，但例程2和例程3都可能使用相同的command1通道向例程1发送消息。

Go中另一个常见的习惯是将回复频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计划：）

Answer 2

GO不可能，我的意思是创建并发频道。

Google Go语言中的并发例程

2 个答案: