如何在 Golang 中控制 Goroutine 数量_Golang 并发任务限流与资源控制策略

func worker(id int, task string, sem chan struct{}) {
    defer func() { <-sem }() // 任务完成释放信号
    fmt.Printf("Worker %d 执行任务: %s\n", id, task)
    time.Sleep(time.Second) // 模拟处理时间
}
<p>func main() {
tasks := []string{"task1", "task2", "task3", "task4", "task5"}
maxConcurrency := 3
sem := make(chan struct{}, maxConcurrency)</p><pre class='brush:php;toolbar:false;'>for i, task := range tasks {
    sem <- struct{}{} // 占用一个槽位，超了会阻塞
    go worker(i+1, task, sem)
}

// 等待所有任务完成（这里简化处理，实际可用 WaitGroup）
time.Sleep(6 * time.Second)

func main() {
    tasks := []string{"task1", "task2", "task3", "task4", "task5"}
    maxConcurrency := 2
    sem := make(chan struct{}, maxConcurrency)
    var wg sync.WaitGroup
<pre class='brush:php;toolbar:false;'>for i, task := range tasks {
    wg.Add(1)
    go func(id int, t string) {
        defer wg.Done()
        sem <- struct{}{}
        defer func() { <-sem }()

        fmt.Printf("Worker %d 处理: %s\n", id, t)
        time.Sleep(time.Second)
    }(i+1, task)
}

wg.Wait() // 等待全部完成
fmt.Println("所有任务结束")

import "golang.org/x/sync/semaphore"
<p>func main() {
tasks := []string{"t1", "t2", "t3", "t4", "t5"}
sem := semaphore.NewWeighted(2) // 最大并发 2
var wg sync.WaitGroup</p><pre class='brush:php;toolbar:false;'>for i, task := range tasks {
    wg.Add(1)
    go func(id int, t string) {
        defer wg.Done()
        sem.Acquire(context.Background(), 1) // 获取许可
        defer sem.Release(1)                // 释放

        fmt.Printf("Worker %d 执行: %s\n", id, t)
        time.Sleep(time.Second)
    }(i+1, task)
}

wg.Wait()

defer func() {
    if r := recover(); r != nil {
        fmt.Printf("Goroutine panic: %v\n", r)
    }
    <-sem
}()