Why Does `time.Sleep` Matter for Concurrent Goroutine Execution in Go?

The Significance of time.sleep in Maintaining Goroutine Execution

In the realm of Go programming, goroutines offer a powerful mechanism for concurrent execution. However, questions arise when certain goroutines fail to run as intended when time.sleep is excluded.

The Code and Observation

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 with time.sleep, this code produces the expected output, where "world" and "hello" are written to the screen interchangeably five times. However, commenting out time.sleep leads to an unexpected result - only "hello" is printed five times.

The Role of time.sleep

The pivotal role of time.sleep lies in providing an opportunity for the goroutine to execute. Without sleep, the goroutine scheduler remains dormant, preventing the say("world") goroutine from gaining control.

The Non-Preemptive Nature of Goroutine Scheduling

Unlike most multitasking operating systems, Go's goroutine scheduler is non-preemptive. This means that goroutines will not be interrupted and yield control unless they explicitly request it.

The Need for Yielding Control

In the absence of time.sleep, the primary goroutine monopolizes control and runs five times without relinquishing it. Once it completes, the program terminates since there are no active goroutines to sustain it.

Conclusion

The inclusion of time.sleep ensures that the secondary goroutine is not starved of execution opportunity. It allows the scheduler to switch control between goroutines, resulting in the intended concurrent behavior. Therefore, understanding time.sleep's role in maintaining goroutine execution is crucial for harnessing the full potential of Go's concurrency features.

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