2017-04-03 09:23
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I am looking at the example from

func main() {
    in := gen(2, 3)

    // Distribute the sq work across two goroutines that both read from in.
    c1 := sq(in)

    // When does this line below execute and what is in `in`?
    c2 := sq(in)

    // Consume the merged output from c1 and c2.
    for n := range merge(c1, c2) {
        fmt.Println(n) // 4 then 9, or 9 then 4

When does c2 := sq(in) run? As what I understand, it executes not when previous line finishes, but instantly as that is a goroutine.

Will c2 receive the next incoming message that is after coming after the message that is received by c1?

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  func main(){
 in:= gen(2  ,3)
 c1:= sq(in)
 //下面的这一行何时执行以及in中的内容 ?
 c2:= sq(in)
为n:= range merge(c1,c2){
 fmt.Println(n)// 4 然后9或9然后4 

c2:= sq(in)何时运行? 据我了解,它不是在上一行结束时执行,而是立即执行。

c2 接收到来的下一条传入消息 在 c1 收到的消息之后吗?

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1条回答 默认 最新

  • doukeyong3746487 2017-04-03 15:51

    Your code does not use goroutines, in order to use go routines you should do something like this:

    q := make(chan type) 
    go sq(in, q)
    go sq(in, q)
    for elem := range q {

    and sq must return the value through a channel

    func sq(in type, q chan type) {
         q <- valueFromIn

    Also you can use WaitGroup to wait for goroutines to finish.

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