如何在 3 秒内打印出这段 Go 代码？

Question

这是 Go 代码 https://www.intervue.io/sandbox-ILSCXZ6RR

func worker() chan int {
   ch := make(chan int)

   go func() {
      time.Sleep(3 * time.Second)
      ch <- 42
   }()

   return ch
}

func main() {
   timeStart := time.Now()

   _, _ = <-worker(), <-worker()

   println(int(time.Since(timeStart).Seconds())) // 3 or 6 ?
}

我怎样才能使它在 3 秒内执行而不是在 6 秒内执行？

Answer 1

需要 6 秒，因为您正在从 worker() 返回的通道接收数据，因此第二个 worker() 无法启动，直到从第一个接收到值，这需要 3 秒。

您正在使用元组赋值。 Spec: Assignments:

The assignment proceeds in two phases. First, the operands of index expressions and pointer indirections (including implicit pointer indirections in selectors) on the left and the expressions on the right are all evaluated in the usual order. Second, the assignments are carried out in left-to-right order.

和Spec: Order of evaluation:

...when evaluating the operands of an expression, assignment, or return statement, all function calls, method calls, and communication operations are evaluated in lexical left-to-right order.

首先启动 2 个 worker，然后从通道接收，因此 goroutine 可以真正运行并发：

ch1, ch2 := worker(), worker()
_, _ = <-ch1, <-ch2

这样，输出将是（在 Go Playground 上尝试）：

Answer 2

您也可以删除频道，因为您没有使用频道的结果，只需使用 sync.WaitGroup

func worker(n int, wg *sync.WaitGroup) {
    defer wg.Done()
    time.Sleep(3 * time.Second)
    fmt.Println("done")
}

func main() {
    timeStart := time.Now()
    var wg sync.WaitGroup

    for i := 0; i <= 1; i++ {
        wg.Add(1)
        go worker(i, &wg)
    }
    wg.Wait()

    fmt.Println(int(time.Since(timeStart).Seconds())) // 3
}

示例代码如下：https://play.golang.org/p/9_Ymx0tcHqS

如何在 3 秒内打印出这段 Go 代码？

How to make this Go code print in 3 seconds?

concurrency

time

channel

go

goroutine