1.5.6 中控制并发执行的Goroutine的最大数目代码示例是否有误

Author: JabinGPCreated Apr 8, 2020Updated Jan 18, 2024

提示：1.5.6 基于Channel的通信中控制并发执行的Goroutine的最大数目代码示例

书中原文

我们可以根据控制Channel的缓存大小来控制并发执行的Goroutine的最大数目, 例如:

var limit = make(chan int, 3)

func main() {
    for _, w := range work {
        go func() {
            limit <- 1
            w()
            <-limit
        }()
    }
    select{}
}

疑惑

代码中没有给出work变量的定义，我个人猜测为函数数组，如下

var work = []func(){
	func() { log.Println("1"); time.Sleep(1 * time.Second) },
	func() { log.Println("2"); time.Sleep(1 * time.Second) },
	func() { log.Println("3"); time.Sleep(1 * time.Second) },
	func() { log.Println("4"); time.Sleep(1 * time.Second) },
	func() { log.Println("5"); time.Sleep(1 * time.Second) },
	func() { log.Println("6"); time.Sleep(1 * time.Second) },
	func() { log.Println("7"); time.Sleep(1 * time.Second) },
	func() { log.Println("8"); time.Sleep(1 * time.Second) },
}

如果work为函数数组，Goroutine在main中的创建并不是同步的，匿名函数中是否应该需要考虑w为闭包引用变量的问题？如定义一个同名局部变量覆盖w来复制一份w的值。

var limit = make(chan int, 3)
var work = []func(){
	func() { log.Println("1"); time.Sleep(1 * time.Second) },
	func() { log.Println("2"); time.Sleep(1 * time.Second) },
	func() { log.Println("3"); time.Sleep(1 * time.Second) },
	func() { log.Println("4"); time.Sleep(1 * time.Second) },
	func() { log.Println("5"); time.Sleep(1 * time.Second) },
	func() { log.Println("6"); time.Sleep(1 * time.Second) },
	func() { log.Println("7"); time.Sleep(1 * time.Second) },
	func() { log.Println("8"); time.Sleep(1 * time.Second) },
}

func main() {
    for _, w := range work {
        w := w
        go func() {
            limit <- 1
            w()
            <-limit
        }()
    }
    select{}
}

Source: chai2010/advanced-go-programming-book

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