如何确保goroutine在main（）之前退出？不使用WaitGroup

From的代码如下：

package main

import (
    "fmt"
    "runtime"
    "time"
)

var signal = make(chan struct{})

func printNumbers() {

    counter := 1
    for {
        select {
        case <-signal:
            fmt.Println("Received signal")
            // do some housekeeping
            return
        default:
            time.Sleep(100 * time.Millisecond)
            counter++
        }
    }
}

func main() {

    go printNumbers()

    fmt.Println("Before: active go-routines", runtime.NumGoroutine())
    time.Sleep(time.Second)
    close(signal)
    //time.Sleep(1 * time.Second) do some work
    fmt.Println("After: active go-routines", runtime.NumGoroutine())
    fmt.Println("Program exited")
}

预期产出为：

Before: active go-routines 2
After: active go-routines 2
Program exited
Received signal

Before: active go-routines 2
After: active go-routines 1   
Program exited
Received signal

主要是活动go例程（之后）输出应为1

---

如何确保只有在其他go例程退出后才能退出go例程？

使用通道等待一个go例程很容易，如下所示。等待多个goroutine时使用WaitGroup

package main

import (
    "fmt"
    "runtime"
    "time"
)

var signal = make(chan struct{})
var done = make(chan struct{})   // closed when goroutine is done.

func printNumbers() {
    defer close(done)
    counter := 1
    for {
        select {
        case <-signal:
            fmt.Println("Received signal")
            // do some housekeeping
            return
        default:
            time.Sleep(100 * time.Millisecond)
            counter++
        }
    }
}

func main() {

    go printNumbers()

    fmt.Println("Before: active go-routines", runtime.NumGoroutine())
    time.Sleep(time.Second)
    close(signal)
    //time.Sleep(1 * time.Second) do some work
    <-done   // wait for the goroutine to be returning.
    fmt.Println("After: active go-routines", runtime.NumGoroutine())
    fmt.Println("Program exited")
}

主程序包
进口(
“fmt”
“运行时”
“时间”
)
var信号=make（chan结构{}）
var done=make（chan struct{}）//在goroutine完成时关闭。
func printNumber（）{
延迟关闭（完成）
计数器：=1
为了{
挑选{
案例