得到"致命的错误:所有的goroutines都睡着了 - 僵局!" 使用sync.WaitGroup时

Question

我试图剥离一组goroutines,然后等待它们全部完成.

import "sync"

func doWork(wg sync.WaitGroup) error {
    defer wg.Done()
    // Do some heavy lifting... request URL's or similar
    return nil
}

func main() {
    var wg sync.WaitGroup
    for i := 0; i < 10; i++ {
        wg.Add(1)
        go doWork(wg)
    }
    wg.Wait()
}

但是,当我运行此代码时,我收到以下错误:

fatal error: all goroutines are asleep - deadlock!

goroutine 16 [semacquire]:
sync.runtime_Semacquire(0xc20818c658)
    /usr/local/Cellar/go/1.3/libexec/src/pkg/runtime/sema.goc:199 +0x30
sync.(*WaitGroup).Wait(0xc2080544e0)
    /usr/local/Cellar/go/1.3/libexec/src/pkg/sync/waitgroup.go:129 +0x14b
main.main()
    /Users/kevin/code/vrusability/scripts/oculus_share_ratings.go:150 +0x398

我很困惑,因为我写的几乎与文档示例演示的完全相同.

Answer 1

您需要将指针传递给WaitGroup,而不是WaitGroup对象.当您传递实际的WaitGroup时,Go会复制该值,并调用Done()该副本.结果是原始WaitGroup将有十个Add并且没有Done,并且WaitGroup的每个副本将具有一个Done(),并且当WaitGroup传递给该函数时有许多Add.

改为传递指针,每个函数都将引用相同的WaitGroup.

import "sync"

func doWork(wg *sync.WaitGroup) error {
    defer wg.Done()
    // Do some heavy lifting... request URL's or similar
    return nil
}

func main() {
    wg := &sync.WaitGroup{}
    for i := 0; i < 10; i++ {
        wg.Add(1)
        go doWork(wg)
    }
}