How to Gracefully Stop a Goroutine in Go Using Signals?

Stopping a Goroutine with Signals

In Go, goroutines are lightweight threads that are used for concurrency. However, there may be scenarios where we need to terminate a goroutine prematurely. This article explores how to achieve this by using signaling mechanisms.

In the provided code, the goal is to stop a goroutine when a specific condition (tooLate) becomes true. However, using a second channel as proposed would indeed block the goroutine when read upon.

Using an Additional Channel with Select

A better approach involves creating a separate channel (tooLate) and using select in the goroutine to monitor both the processing channel (proCh) and the tooLate channel. When tooLate receives a signal, the goroutine can terminate gracefully.

<code class="go">package main

import "fmt"
import "time"

func main() {
    tooLate := make(chan struct{})
    proCh := make(chan string)

    go func() {
        for {
            fmt.Println("working")
            time.Sleep(1 * time.Second)
            select {
            case <p>In this updated code, when the timer expires (indicating it's too late), the tooLate channel is closed, sending a signal to the goroutine. The goroutine receives this signal and terminates accordingly.</p>
<p><strong>Using sync.Cond</strong></p>
<p>Another option to consider is using sync.Cond, which provides more granular control over signaling. However, it requires a bit more setup and synchronization logic.</p></code>

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