英文:
Why does this WaitGroup sometimes not wait for all goroutines?
问题
下面的代码有时会输出2。为什么等待组（wait group）没有等待所有的goroutine完成？
type Scratch struct {//sync.RWMutexItch []int}func (s *Scratch) GoScratch(done chan bool, j int) error {var ws sync.WaitGroupif len(s.Itch) == 0 {s.Rash = make([]int, 0)}for i := 0; i < j; i++ {ws.Add(1)go func (i int) {defer ws.Done()s.Rash = append(s.Rash, i)}(i)}ws.Wait()done <- truereturn nil}func main() {done := make(chan bool, 3)s := &Scratch{}err := s.GoScratch(done, 3)if err != nil {log.Println("Error: ", err)}<-donelog.Println("Length: ", len(s.Rash))}
奇怪的是，我无法通过main函数输出2，但是当我使用一个测试用例时，有时会输出2。
英文:
The code below outputs 2 sometimes. Why isn't the wait group waiting for all the goroutines to complete ?
type Scratch struct {//sync.RWMutexItch []int}func (s *Scratch) GoScratch(done chan bool, j int) error {var ws sync.WaitGroupif len(s.Itch) == 0 {s.Rash = make([]int, 0)}for i := 0; i < j; i++ {ws.Add(1)go func (i int) {defer ws.Done()s.Rash = append(s.Rash, i)}(i)}ws.Wait()done<- truereturn nil}func main() {done := make(chan bool, 3)s := &Scratch{}err := s.GoScratch(done, 3)if err != nil {log.Println("Error:%v",err)}<-donelog.Println("Length: ", len(s.Rash))}`
Strangely i can't get it to output 2 with a main function but when I use a test case it outputs 2 sometimes.
答案1
得分: 10
你的代码中存在竞态条件。它就在这里：
go func (i int) {defer ws.Done()// 在 s.Rash 访问上存在竞态条件s.Rash = append(s.Rash, i)}(i)
由于所有的 goroutine 都同时访问 s.Rash，这可能导致切片的更新被覆盖。尝试使用 sync.Mutex 运行相同的代码来防止这种情况发生：
// 创建一个全局互斥锁var mutex = &sync.Mutex{}// 使用互斥锁来防止竞态条件go func (i int) {defer ws.Done()defer mutex.Unlock() // 确保互斥锁解锁// 在访问资源之前锁定它mutex.Lock()s.Rash = append(s.Rash, i)}(i)
英文:
There is a race condition in your code. It is right here:
go func (i int) {defer ws.Done()// race condition on s.Rash accesss.Rash = append(s.Rash, i)}(i)
Since all the goroutines access s.Rash concurrently, this may cause the slice updates to be overwritten. Try running the same code with sync.Mutex locking to prevent this:
// create a global mutexvar mutex = &sync.Mutex{}// use mutex to prevent race conditiongo func (i int) {defer ws.Done()defer mutex.Unlock() // ensure that mutex unlocks// Lock the resource before accessing itmutex.Lock()s.Rash = append(s.Rash, i)}(i)
答案2
得分: 6
如果你使用竞争检测器运行你的代码：
go test -race .
你会发现在切片s.Rash上存在竞争条件。
英文:
If you run your code with race detector
go test -race .
you will find out race condition on slice s.Rash.
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