最经典的处理方式：
在启动goroutine的时候，传递一个额外的chan型参数，用来接收退出信号，代码如下
func worker(name string, stopchan chan struct{}) {
for {
select {
case <-stopchan:
fmt.Println("receive a stop signal, ", name)
return
default:
fmt.Println("I am worker ", name)
time.Sleep( * time.Second)
}
}
}
在main函数中应该如何发送stop信号呢？
func main() {
stopCh := make(chan struct{})
go worker("a", stopCh)
time.Sleep( * time.Second)
stopCh <- struct{}{}
time.Sleep( * time.Second)
}
输出：
I am worker a
I am worker a
receive a stop signal, a Process finished with exit code
ok，从输出可以看出name为a的这个woker在收到信号之后退出了，过了2s后主函数退出
当我们又2个goroutine的时候情况如何呢？
func main() {
stopCh := make(chan struct{})
go worker("a", stopCh)
go worker("b", stopCh)
time.Sleep( * time.Second)
stopCh <- struct{}{}
time.Sleep( * time.Second)
}
输出：
I am worker a
I am worker b
I am worker b
I am worker a
receive a stop signal, a
I am worker b
I am worker b
I am worker b
也就是说a退出了，b没有退出，因为stopCh <- struct{}{}只发送一个信号，被a接收了，b不受影响
如果想让2个goroutine同时退出，需要这样写：
close(stopCh)
再看下输出：
I am worker a
I am worker b
I am worker a
receive a stop signal, b
receive a stop signal, a Process finished with exit code
已经全部退出了