有缓冲的channel

package main

import (
	"fmt"
	"math/rand"
	"sync"
	"time"
)

const (
	numberGoroutines = 4
	taskLoad         = 10
)

var wg sync.WaitGroup

func init() {
	rand.Seed(time.Now().Unix())
}

func worker(tasks chan string, worker int) {
	defer wg.Done()
	for {
		task, ok := <-tasks
		if !ok {
			fmt.Printf("Workerr : %d : Shutting Down\n", worker)
			return
		}
		fmt.Printf("Worker : %d : Started work %s\n", worker, task)
		sleep := rand.Int63n(100)
		time.Sleep(time.Duration(sleep) * time.Microsecond)
		fmt.Printf("Worker : %d : Completed %s \n", worker, task)
	}
}

func main() {
	tasks := make(chan string, 10)
	wg.Add(numberGoroutines)
	// 开启消费者
	for gr := 1; gr <= numberGoroutines; gr++ {
		go worker(tasks, gr)
	}
	// 开启生产者
	for post := 1; post <= taskLoad; post++ {
		tasks <- fmt.Sprintf("Task : %d", post)
	}
	close(tasks)
	wg.Wait()
}

有一个疑问就是，如果tasks这个channel换成无缓冲的，结果看上去有点奇怪