goroutines执行顺序

Question

我试图理解这段代码，不确定为什么第二遍在第一遍之前执行。如果有人真的可以帮助我，那就太好了！

func sum(a []int, c chan int) {
   fmt.Println("summing: ", a)
   total := 0
   for _, v := range a {
      total += v
   }
   //fmt.Println("send to c",total)
   c <- total  // send total to c
}
func main() {
    //a := []int{7, 2, 8,134,23,23,1,23,1234,143, -9, 4, 0, 1234}

    c := make(chan int)

    go sum([]int{1,2,3}, c)
    go sum([]int{4,5,6}, c)

    x := <-c
    fmt.Println(x)
    x = <-c
    fmt.Println(x)
}

输出：

summing:  [4 5 6]
15
summing:  [1 2 3]
6

Answer 1

您没有任何内容可以显式同步两个goroutine的顺序。如果运行足够的时间，您将看到调用以fmt.Println不同的顺序进行打印。当执行goroutine时，由于它们是并发操作，因此无法保证它们将何时执行和/或完成。您需要使用各种标准库程序包或通道本身来同步并发运行的goroutine的执行。

例如（通过利用渠道的封锁性，您可以执行以下操作）：

func main() {

    c := make(chan int)

    go sum([]int{1, 2, 3}, c)

    //use the channel to block until it receives a send
    x := <-c
    fmt.Println(x)

    //then execute the next routine
    go sum([]int{4, 5, 6}, c)

    x = <-c
    fmt.Println(x)
}

另一个示例（显然不那么实用，但是在这里查看其他常见的go同步功能），您可以引入一个等待组和一个通道范围：

func sum(a []int, c chan int, wg *sync.WaitGroup) {
    defer wg.Done()
    fmt.Println("summing: ", a)
    total := 0
    for _, v := range a {
        total += v
    }
    //fmt.Println("send to c",total)
    c <- total // send total to c
}

func main() {

    c := make(chan int)
    wg := new(sync.WaitGroup)

    //concurrently call the concurrent calls to sum, allowing execution to continue to the range of the channel 
    go func() {
        //increment the wait group, and pass it to the sum func to decrement it when it is complete
        wg.Add(1)
        go sum([]int{1, 2, 3}, c, wg)
        //wait for the above call to sum to complete
        wg.Wait()
        //and repeat...
        wg.Add(1)
        go sum([]int{4, 5, 6}, c, wg)
        wg.Wait()
        //all calls are complete, close the channel to allow the program to exit cleanly 
        close(c)
    }()

    //range of the channel
    for theSum := range c {
        x := theSum
        fmt.Println(x)
    }

}