Why Does an Infinite Go Routine Block Other Goroutines from Sending to Channels?

Go Routine Blocking Others: A Deeper Explanation

In Go, the following code exhibits an unusual behavior where a goroutine with an infinite loop seemingly prevents another goroutine's message from reaching the intended channel:

func main() {
   timeout := make(chan int)
   go func() {
      time.Sleep(time.Second)
      timeout <- 1
    }()

    res := make(chan int)
    go func() {
        for {
        }
        res <- 1
    }()
    select {
        case <-timeout:
            fmt.Println("timeout")
        case <-res:
            fmt.Println("res")
    }
}

Instead of terminating after one second, the program enters an infinite loop. Why does this happen?

Understanding Cooperative Scheduling in Go

The answer lies in Go's use of cooperative scheduling for goroutines. Goroutines yield control to the scheduler under certain conditions, including:

• Unbuffered channel send/receive operations
• System calls (e.g., file/network I/O)
• Memory allocation
• time.Sleep() calls
• runtime.Gosched() calls

Since the infinite loop in the first goroutine never yields, it prevents other goroutines from running and sending messages to channels. This includes the timeout channel, which is waiting for a message that will never arrive.

Potential Solutions

Although cooperative scheduling can lead to such situations, there are potential solutions:

• Increasing GOMAXPROCS: This environment variable allows multiple threads to execute goroutines concurrently, reducing the likelihood of one goroutine blocking others.
• Using a Preemptive Scheduler (Future Goal): Go language developers aim to implement a preemptive scheduler, which would forcibly switch between goroutines, eliminating the issue of blocking.
• Manual Yielding: The runtime.Gosched() function allows goroutines to manually yield control to the scheduler, breaking out of infinite loops that might block others.

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