go语言中带缓冲区的channel的使用

一.本文主要针对以下问题:

1.带缓冲区的channel

2.在两个goroutine之间进行消息的交互

3.偶尔会有消息到达,有时又阻塞于没有消息到达

4.两个goroutine一直存在

设计思路,首先要保证channel构造完成前,发送和接收的消息都没有使用channel,不然会导致channel阻塞

二.下面是错误的代码:

1.发送端:

package ioproc

import (
   "fmt"
   "time"
)

var ChT = make([]chan int,4)
func Send(){
   fmt.Println("Send entry.....")

   for i:=0;i<4;i++{
      go func(i int){
         var temp = 0
         ChT[i] = make(chan int,5000)

         for {
            time.Sleep(time.Millisecond*1000)
            temp++
            ChT[i] <- temp
            fmt.Println("Send temp ok,",i,"Send")
         }

      }(i)
   }

}

  


代码解释:发送函数部分,1.声明了一个全局的channel数组变量,ChT[]
            2.定义send函数:使用for循环,在每个for循环开启一个goroutine,并将ChT[i]进行make为缓冲区大小是5000的channel
            3.一直执行for循环,往ChT[i]中进行添加数据

2.接收端
package utility

import (
   "SIPLBS/ioproc"
   "fmt"
)

func Recv(){
   for{
      select{
      case tNumber := <- ioproc.ChT[0]:
         fmt.Println("the ",0,"routine is ",tNumber)

      case tNumber := <- ioproc.ChT[1]:
         fmt.Println("the ",1,"routine is ",tNumber)
      case tNumber := <- ioproc.ChT[2]:
         fmt.Println("the ",2,"routine is ",tNumber)

      case tNumber := <- ioproc.ChT[3]:
         fmt.Println("the ",3,"routine is ",tNumber)
      }
   }
}

  


代码解释:接收函数部分:1.定义函数,并在函数中一直for循环,通过select一直接收ChT[i]中的数据

3.main包函数:
func init(){
   fmt.Println("init entry......")

   go ioproc2.Send()

   go ioproc.Recv()
}

 

4.打印结果:

init entry......
Send entry.....
sendsip begin...
Send temp ok, 1 Send
Send temp ok, 0 Send
Send temp ok, 3 Send
Send temp ok, 2 Send
Send temp ok, 0 Send
Send temp ok, 1 Send
Send temp ok, 3 Send
Send temp ok, 2 Send
Send temp ok, 2 Send

可以看到一直有消息往channel中发送,但是recv的GoRoutine一直未收到消息


三.修改之后的代码

1.发送端
import (
	"fmt"
	"time"
)

var ChT = make([]chan int,4)

func init(){
	fmt.Println("init entry......")

	for i := 0 ; i < 4; i++{
		ChT[i] = make(chan int,5000)
	}
}

func Send(){
	fmt.Println("Send entry.....")

	for i:=0;i<4;i++{
		go func(i int){
			var temp = 0
			for {
				time.Sleep(time.Millisecond*1000)
				temp++
				ChT[i] <- temp
				fmt.Println("Send temp ok,",i,"Send")
			}

		}(i)
	}
}

2.接收端

package utility

import (
	"SIPLBS/ioproc"
	"fmt"
)

func Recv(){
	for{
		select{
		case tNumber := <- ioproc.ChT[0]:
			fmt.Println("the ",0,"routine is ",tNumber)

		case tNumber := <- ioproc.ChT[1]:
			fmt.Println("the ",1,"routine is ",tNumber)
		case tNumber := <- ioproc.ChT[2]:
			fmt.Println("the ",2,"routine is ",tNumber)

		case tNumber := <- ioproc.ChT[3]:
			fmt.Println("the ",3,"routine is ",tNumber)
		}
	}
}

  3.main包程序

func main() {

	go ioproc2.Send()

	go utility.Recv()
}

 代码解释:目的是将channel放在send和recv两个goroutine启动之前已经准备好,所以将channel的初始化部分放到了先于main执行的init函数中,然后在main函数中起send和recv协程。

4.结果打印:

init entry......
Send entry.....
Send temp ok, 1 Send
Send temp ok, 3 Send
Send temp ok, 2 Send
the 3 routine is 1
the 1 routine is 1
the 2 routine is 1
Send temp ok, 0 Send
the 0 routine is 1

 

还可以改进代码,将三个函数都封装到一个类中进行处理,在类进行初始化的时候,可以把channel进行初始化,

然后开启goroutine,这样就可以确保发送和接收的goroutine是在channel准备好的情况下进行执行的。

 





 

posted @ 2020-03-25 15:02  知更之始,博之于涵  阅读(993)  评论(0编辑  收藏  举报