douqiao6015 2019-08-29 00:53
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如何两次解锁互斥锁

Is it safe to unlock a mutex twice? My code:

var m sync.RWMutex = sync.RWMutex{}

func Read() {
    m.RLock()
    defer m.RUnlock()

    // Do something that needs lock
    err := SomeFunction1()
    if err != nil {
        return
    }

    m.RUnlock()

    // Do something that does not need lock
    SomeFunction2()

}

I need defer m.RUnlock() for the case SomeFunction1() returns error. But when SomeFunction1() returns without error, m will be unlocked twice by m.RUnlock() and defer m.RUnlock().

Is it safe to unlock the mutex twice? If not, how should I fix my code?

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  • doulongti5932 2019-08-29 01:14
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    Is it safe for unlocking a mutex twice?

    Nope, you shouldn't unlock the mutex twice. It's a run-time error according to docs.

    RUnlock undoes a single RLock call; it does not affect other simultaneous readers. It is a run-time error if rw is not locked for reading on entry to RUnlock.


    If not, how should I fix my code?

    I would recommend to keep the defer but only m.RUnlock() in case of error. This can easily scale in case you add more function calls between SomeFunction1() and SomeFunction2().

    func Read() {
        var err error
        m.RLock()
        defer func() {
            if err != nil {
                m.RUnlock()
            }
        }()
    
    
        // Do something that needs lock
        err = SomeFunction1()
        if err != nil {
            return
        }
    
        m.RUnlock()
    
        // Do something that does not need lock
        SomeFunction2()
    }
    

    Try it on Go Playground!

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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