Go语言快速入门笔记(二):并发编程部分

1.协程是轻量级的线程的理解

(1)java里的Thread默认为1M,Groutine的Stack初始化为2K
(2)一个thread对应一个kernel space,groutine则是多对多

2.第一个简单的多线程案列

 for i := 0; i < 10; i++ {
 go func(i int) {
 println(i)
 }(i)
 }
}

3.如何做到多线程累加同一数据的安全性

func TestGroutine1(t *testing.T) {
 count := 0
 mutex := sync.Mutex{}
 for i := 0; i < 10; i++ {
 go func() {
 defer func() {
 mutex.Unlock()
 }()
 mutex.Lock()
 count++
 println(count)
 }()
 }
}

4.如何打印完10个数主线程再输出Done?

func TestGroutine1(t *testing.T) {
 count := 0
 mutex := sync.Mutex{}
 wg := sync.WaitGroup{}
 for i := 0; i < 10; i++ {
 wg.Add(1)
 go func() {
 defer func() {
 mutex.Unlock()
 }()
 mutex.Lock()
 count++
 println(count)
 wg.Done()
 }()
 }
 wg.Wait()
 println("done")
}

5.CSP并发机制

结合一个前端开发的场景,使用ajax异步去数据
会阻塞在取数据的channel上

func AsyncAJAX() chan string {
 channel := make(chan string)
 data := "Hello world"
 go func() {
 time.Sleep(time.Second * 10)
 println("ajax执行完毕")
 channel <- data
 }()
 return channel
}
func TestCSP(t *testing.T) {
 println("Ajax发送请求")
 c := AsyncAJAX()
 println("渲染")
 println("取数据",<-c)
}

6.select多路选择做超时控制

func TestCSP(t *testing.T) {
 println("Ajax发送请求")
 c := AsyncAJAX()
 select {
 case ret := <-c:
 t.Log(ret)
 case <- time.After(time.Second * 5):
 t.Log("time out")
 }
}

7.close掉channel会迅速接到0值

8.带buffer的话就可以堆积

func dataSend(c chan int) {
 go func() {
 for i := 0;;i++ {
 time.Sleep(time.Second * 1)
 println("发送第", i)
 c <- i
 }}()
}
func dataRecv(c chan int) {
 go func() {
 for {
 time.Sleep(time.Second * 5)
 println("接收第", <- c)
 }}()
}
func TestMore(t *testing.T) {
 c := make(chan int, 5)
 dataSend(c)
 dataRecv(c)
 time.Sleep(time.Second * 100)
}

9.任意一个任务完成

func anyComplete() int {
 channel := make(chan int)
 for i := 0; i < 10; i++ {
 go func(i int) {
 time.Sleep(time.Duration(rand.Int31n(10)))
 channel <- i
 }(i)
 }
 return <-channel
}
func TestAnyComplete(t *testing.T) {
 t.Log("result is from ", anyComplete())
}

10.所有任务都执行完成

func TestAllComplete(t *testing.T) {
 channel := make(chan int)
 for i := 0; i < 10; i++ {
 go func(i int) {
 time.Sleep(time.Duration(rand.Int31n(10)))
 channel <- i
 }(i)
 }
 sum := 0
 for i := 0; i < 10; i++ {
 sum += <- channel
 }
 println(sum)
}