Deadlock between goroutines

倖福魔咒の 提交于 2019-12-24 08:22:50

问题


I'm new to Go. When I comment out the second goroutine, there is a fatal error. I don't understand what causes this error to occur. Can you explain it to me?

package main

import (
    "fmt"
    "time"
)

func main() {
    ch := make(chan int)
    go func() { 
        for i := 0; i < 10; i++ {
            ch <- i
        }
    } ()
    // go func() { 
        for {
            if num, ok := <-ch; !ok {
                break
            } else {
                fmt.Printf("%d\n", num)
            }
        }
    // } ()
    time.Sleep(2 * time.Second)
    close(ch)
}

This prints the following code:

0
1
2
3
4
5
6
7
8
9
fatal error: all goroutines are asleep - deadlock!

goroutine 1 [chan receive]:
main.main()
    /tmp/sandbox169127128/main.go:17 +0xa0

Program exited.

回答1:


The receiving for loop blocks on receive from ch after receiving all values from the sending goroutine. The runtime detects that the program is stuck and panics.

The fix is to close the channel after sending all values:

go func() { 
    for i := 0; i < 10; i++ {
        ch <- i
    }
    close(ch)
} ()

Receive on the closed channel yields the value 0, false. The receiving for loop breaks on the false value.

Remove close(ch) from the end of the program.

Run it on the playground.




回答2:


Because you're not closing the channel before the first goroutine exits. The below code should work.

package main

import (
    "fmt"
    "time"
)

func main() {
    ch := make(chan int)
    go func() { 
        for i := 0; i < 10; i++ {
            ch <- i
        }
        close(ch)
    } ()
    //go func() { 
        for {
            if num, ok := <-ch; !ok {
                break
            } else {
                fmt.Printf("%d\n", num)
            }
        }
    //} ()
    time.Sleep(2 * time.Second)
}

Try it out here: https://play.golang.org/p/OdxNqbaZmj



来源:https://stackoverflow.com/questions/43297474/deadlock-between-goroutines

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