Breaking News: Grepper is joining You.com. Read the official announcement!

golang channel handling

Restu Wahyu Saputra answered on August 21, 2023 Popularity 3/10 Helpfulness 3/10

answer golang channel handling

golang channel handling

Comment

// You can edit this code!
// Click here and start typing.
package main

import (
	"errors"
	"fmt"
	"log"
	"sync"
	"time"
)

var (
	wg           *sync.WaitGroup = &sync.WaitGroup{}
	mutex        *sync.RWMutex   = &sync.RWMutex{}
	deliveryChan chan []byte     = make(chan []byte, 1)
	res          []byte          = []byte{}
)

func Execute() ([]byte, error) {
	// deliveryChan <- []byte("Hello World")
	<-time.After(time.Duration(time.Second * 1))

	wg.Add(2)
	go func() {
		wg.Done()

		mutex.Lock()
		defer mutex.Unlock()

		res = <-deliveryChan
	}()

	go func() {
		wg.Done()

		time.AfterFunc(time.Second*5, func() {
			deliveryChan <- nil
			res = <-deliveryChan
		})
	}()
	wg.Wait()

	mutex.RLock()
	defer mutex.RUnlock()

	if len(res) <= 0 {
		return nil, errors.New("Empty Response")
	}

	return res, nil
}

func main() {
	res, err := Execute()

	if err != nil {
		log.Fatal(err.Error())
	}

	fmt.Println("Response: ", string(res))
}

Popularity 3/10 Helpfulness 3/10 Language go

Source: Grepper

Tags: channel go

Link to this answer
Share Copy Link

Contributed on Aug 21 2023

Restu Wahyu Saputra

0 Answers Avg Quality 2/10

Continue with Google

By continuing, I agree that I have read and agree to Greppers's Terms of Service and Privacy Policy.

golang channel handling

Contents

More Related Answers

golang channel handling

Grepper

Documentation

Social

Legal

Contact

Oops, You will need to install Grepper and log-in to perform this action.