How Can a Goroutine with an Infinite Loop Block Other Goroutines in Go?

Goroutines Blocking Others

In Go, a goroutine is a lightweight thread of execution. Normally, goroutines can run concurrently without blocking each other. However, under certain circumstances, one goroutine can block the execution of others.

Consider the following code:

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")
    }
}

In this example, a goroutine is started with an infinite loop. Normally, this goroutine should not prevent other goroutines from executing. However, it appears that the first goroutine is somehow blocking the second goroutine from sending to the timeout channel.

Explanation

This behavior is due to the way that goroutines are scheduled in Go. Cooperative scheduling is used, which means that goroutines must voluntarily yield execution to the scheduler in order for another goroutine to run. In the provided code, the goroutine with the infinite loop is never yielding, which prevents the other goroutine from making progress.

Goroutines typically yield under the following conditions:

• Unbuffered channel send/receive
• System calls (e.g., file/network reads/writes)
• Memory allocation
• Time.Sleep() is called
• runtime.Gosched() is called

In this case, none of these conditions are met, so the goroutine with the infinite loop continues running indefinitely.

Solutions

To solve this issue, there are several possible solutions:

• Use a preemptive scheduler. This is a work in progress, but it would allow goroutines to be interrupted at any time, regardless of whether they are yielding.
• Use unbuffered channels. When sending to an unbuffered channel, the sender will block until another goroutine is ready to receive.
• Use a timer. The time.After() function can be used to create a channel that will receive after a specified duration. By sending to this channel, you can force the goroutine to yield.
• Use runtime.Gosched() within the infinite loop. This will explicitly yield execution to the scheduler.

The best solution for your specific application will depend on the nature of the goroutines and the desired behavior.

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