dooso0594 2019-06-24 20:07
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使用for循环遍历通道时获取Goroutine死锁

I am tring to practise goroutine and channels, where I am facing issue where I am calling gorouting and passing channel. The goroutine will push data into into channel and then main thread will print the elements.

I have used for loop to print the contents but getting.

fatal error: all goroutines are asleep - deadlock!

2
1
goroutine 1 [chan receive]:
main.main()
package main

import "fmt"


func smallThread(a int, c chan int) {
    c <- a
}

func main() {
    c := make(chan int)
    go smallThread(1, c)
    go smallThread(2, c)
    for {
        fmt.Println(<-c)
    }
}

EDIT: Using waitgroup:

func smallThread(a int, c chan int, w *sync.WaitGroup) {
    c <- a
    w.Done()
}

func main() {
    c := make(chan int)
    var w sync.WaitGroup
    w.Add(2)
    go smallThread(1, c, &w)
    go smallThread(2, c, &w)
    //w.Wait()
    for i := range c {
        fmt.Println(i)
    }
    w.Wait()
}

EDIT2: Working Code

func smallThread(a int, c chan int, w *sync.WaitGroup) {
    //defer w.Done()
    c <- a
    w.Done()
}

func main() {
    c := make(chan int)
    var w sync.WaitGroup
    w.Add(1)
    go smallThread(1, c, &w)
    w.Add(1)
    go smallThread(2, c, &w)
    go func(c chan int) {
        for i := range c {
            fmt.Println(i)
        }
    }(c)
    w.Wait()
}
  • 写回答

2条回答 默认 最新

  • dongxifu5009 2019-06-24 20:38
    关注

    When the goroutines are done, close the channel to indicate that no more values will be added. The for loop will break when after all values are received.

    c := make(chan int)
    var w sync.WaitGroup
    w.Add(2)
    go smallThread(1, c, &w)
    go smallThread(2, c, &w)
    go func() {
        w.Wait()
        close(c)
    }()
    
    for i := range c {
        fmt.Println(i)
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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