Go-waitGroup的那些事

Add() 和 Wait() 最好不要放在不同的 goroutine, 避免并发执行.同一个 once 实例调用多次 Do 方法, 只有一个函数会被调用, 但是其他的函数会阻塞, 等待 Unlock.

上代码, 自己品:

package main

import (
	"fmt"
	"sync"
)

func main() {
	var wg sync.WaitGroup
	wg.Add(2)
	go func() {
		defer wg.Done()
		fmt.Println(1)
	}()
	go func() {
		defer wg.Done()
		fmt.Println(2)
	}()
	wg.Wait()
	fmt.Println("Done")
}

package main

import (
	"fmt"
	"sync"
	"time"
)

func main() {
	var wg sync.WaitGroup
	wg.Add(1)
	go func() {
		//defer wg.Done()
		fmt.Println(1)
		wg.Done()
		wg.Add(1)
		time.Sleep(time.Second * 1)
		wg.Done()
	}()
	wg.Wait() // panic: sync: WaitGroup is reused before previous Wait has returned
	fmt.Println("Done")
}

package main

import (
	"fmt"
	"sync"
	"time"
)

func main() {
	var wg sync.WaitGroup

	times := 10
	batch := 3

	for i := 0; i < times; i++ {
		wg.Add(batch)
		for j := 0; j < batch; j++ {
			go func(i, j int) {
				defer wg.Done()
				time.Sleep(time.Second)
				fmt.Println(i, j)
			}(i, j)
		}
		wg.Wait()
	}

	fmt.Println("Done")
}

package main

import (
	"fmt"
	"sync"
)

func main() {
	var once sync.once
	once.Do(
		func() {
			fmt.Println(1)
		},
	)
	once.Do(
		func() {
			fmt.Println(2)
		},
	)
	fmt.Println("Done")
}

package main

import (
	"fmt"
	"sync"
)

func main() {
	var once sync.once

	go func() {
		once.Do(
			func() {
				fmt.Println(1)
			},
		)
	}()

	go func() {
		once.Do(
			func() {
				fmt.Println(2)
			},
		)
	}()

	fmt.Println("Done")
}

package main

import (
	"fmt"
	"sync"
	"time"
)

func main() {
	var once sync.once

	go func() {
		// 会阻塞
		once.Do(
			func() {
				fmt.Println(2)
			},
		)
		fmt.Println(2, "done")
	}()
	once.Do(
		func() {
			time.Sleep(time.Second * 3)
			fmt.Println(1)
		},
	)
	fmt.Println("Done")

	time.Sleep(time.Second * 5)
}