How Does `time.Sleep()` Prevent Goroutine Starvation in Go?

Understanding the Role of time.sleep in Goroutine Execution

In Go, goroutines are lightweight concurrently executing threads. While combining goroutines with channel communication provides robust concurrency primitives, certain scenarios call for the usage of blocking functions like time.sleep to prevent starvation and ensure fairness in execution.

Consider the following code snippet from the Go tutorial:

package main

import (
    "fmt"
    "time"
)

func say(s string) {
    for i := 0; i < 5; i++ {
        time.Sleep(100 * time.Millisecond)
        fmt.Println(s)
    }
}

func main() {
    go say("world")
    say("hello")
}

When executed, this code prints "hello" and "world" interchangeably on the screen five times. However, if time.Sleep is commented out, only "hello" will be printed. This behavior highlights the significance of time.Sleep in goroutine execution.

The Go runtime scheduler is not preemptive. This means goroutines must voluntarily relinquish control to allow other goroutines to execute. One way to release control is to use time.Sleep.

In the code snippet above, if time.Sleep is removed, the "hello" goroutine will run for five iterations without giving the "world" goroutine any opportunity to execute. As a result, when the "hello" goroutine finishes, the program terminates due to the absence of any other goroutines keeping it alive.

Therefore, time.Sleep plays a crucial role in ensuring fairness in goroutine scheduling. By introducing short delays into the execution of specific goroutines, it allows other goroutines to be executed and prevents starvation or complete starvation of other goroutines.

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