go RWMutex 的实现

Overview

go 里面的 rwlock 是 write preferred 的,可以避免写锁饥饿。

读锁和写锁按照先来后到的规则持有锁,一旦有协程持有了写锁,后面的协程只能在写锁被释放后才能得到读锁。

同样,一旦有 >= 1 个协程写到了读锁,只有等这些读锁全部释放后,后面的协程才能拿到写锁。

下面了解一下 Go 的 RWMutex 是如何实现的吧,下面的代码取自 go1.17.2/src/sync/rwmutex.go,并删减了 race 相关的代码。

PS: rwmutex 的代码挺短的,其实读源码也没那么可怕...

RWMutex 的结构

RWMutex 总体上是通过: 普通锁和条件变量来实现的

type RWMutex struct {
	w           Mutex  // held if there are pending writers
	writerSem   uint32 // semaphore for writers to wait for completing readers
	readerSem   uint32 // semaphore for readers to wait for completing writers
	readerCount int32  // number of pending readers
	readerWait  int32  // number of departing readers
}
Lock
func (rw *RWMutex) Lock() {
	// First, resolve competition with other writers.
	rw.w.Lock()
	// Announce to readers there is a pending writer.
	r := atomic.AddInt32(&rw.readerCount, -rwmutexMaxReaders) + rwmutexMaxReaders
	// Wait for active readers.
	if r != 0 && atomic.AddInt32(&rw.readerWait, r) != 0 {
		runtime_SemacquireMutex(&rw.writerSem, false, 0)
	}
}
Unlock
const rwmutexMaxReaders = 1 << 30

func (rw *RWMutex) Unlock() {
	// Announce to readers there is no active writer.
	r := atomic.AddInt32(&rw.readerCount, rwmutexMaxReaders)
	// Unblock blocked readers, if any.
	for i := 0; i < int(r); i++ {
		runtime_Semrelease(&rw.readerSem, false, 0)
	}
	// Allow other writers to proceed.
	rw.w.Unlock()
}
RLock
func (rw *RWMutex) RLock() {
	if atomic.AddInt32(&rw.readerCount, 1) < 0 {
		// A writer is pending, wait for it.
		runtime_SemacquireMutex(&rw.readerSem, false, 0)
	}
}
RUnlock
func (rw *RWMutex) RUnlock() {
	if r := atomic.AddInt32(&rw.readerCount, -1); r < 0 {
		// Outlined slow-path to allow the fast-path to be inlined
		rw.rUnlockSlow(r)
	}
}
func (rw *RWMutex) rUnlockSlow(r int32) {
	// A writer is pending.
	if atomic.AddInt32(&rw.readerWait, -1) == 0 {
		// The last reader unblocks the writer.
		runtime_Semrelease(&rw.writerSem, false, 1)
	}
}
Q1: 多个协程并发拿读锁,如何保证这些读锁协程都不会被阻塞?
func (rw *RWMutex) RLock() {
	if atomic.AddInt32(&rw.readerCount, 1) < 0 {
		// A writer is pending, wait for it.
		runtime_SemacquireMutex(&rw.readerSem, false, 0)
	}
}

拿读锁时,仅仅会增加 readerCount,因此读锁之间是可以正常并发的

Q2: 多个协程并发拿写锁,如何保证只会有一个协程拿到写锁?
func (rw *RWMutex) Lock() {
	// First, resolve competition with other writers.
	rw.w.Lock()
	// Announce to readers there is a pending writer.
	r := atomic.AddInt32(&rw.readerCount, -rwmutexMaxReaders) + rwmutexMaxReaders
	// Wait for active readers.
	if r != 0 && atomic.AddInt32(&rw.readerWait, r) != 0 {
		runtime_SemacquireMutex(&rw.writerSem, false, 0)
	}
}

拿写锁时,会获取 w.Lock,自然能保证同一时间只会有一把写锁

Q3: 在读锁被拿到的情况下,新协程拿写锁,如果保证写锁现成会被阻塞?
func (rw *RWMutex) Lock() {
	// First, resolve competition with other writers.
	rw.w.Lock()
	// Announce to readers there is a pending writer.
	r := atomic.AddInt32(&rw.readerCount, -rwmutexMaxReaders) + rwmutexMaxReaders
	// Wait for active readers.
	if r != 0 && atomic.AddInt32(&rw.readerWait, r) != 0 {
		runtime_SemacquireMutex(&rw.writerSem, false, 0)
	}
}

假设此时有 5 个协程拿到读锁,则 readerCount = 5,假设 rwmutexMaxReaders = 100。

此时有一个新的协程 w1 想要拿写锁。

在执行

r := atomic.AddInt32(&rw.readerCount, -rwmutexMaxReaders) + rwmutexMaxReaders

后, rw.readerCount = -95,r = 5。

在执行

atomic.AddInt32(&rw.readerWait, r)

后,rw.readerWait = 5。

readerWait 记录了在获取写锁的这一瞬间有多少个协程持有读锁。这一瞬间之后,就算有新的协程尝试获取读锁,也只会增加 readerCount ,而不会动到 readerWait。

之后执行 runtime_SemacquireMutex() 睡在了 writerSem 这个信号量上面。

Q4: 在读锁被拿到的情况下,新协程拿写锁被阻塞,当旧有的读锁协程全部释放,如何唤醒等待的写锁协程
func (rw *RWMutex) RUnlock() {
	if r := atomic.AddInt32(&rw.readerCount, -1); r < 0 {
		// Outlined slow-path to allow the fast-path to be inlined
		rw.rUnlockSlow(r)
	}
}
func (rw *RWMutex) rUnlockSlow(r int32) {
	// A writer is pending.
	if atomic.AddInt32(&rw.readerWait, -1) == 0 {
		// The last reader unblocks the writer.
		runtime_Semrelease(&rw.writerSem, false, 1)
	}
}

继续上一步的场景,每当执行 RUnlock 时,readerCount 都会减去1。当 readerCount 为负数时,意味着有协程正在持有或者正在等待持有写锁。

之前的五个读协程中的四个,每次 RUnlock() 之后,readerCount = -95 - 4 = -99,readerWait = 5 - 4 = 1。

当最后一个读协程调用 RUnlock() 之后,readerCount 变成了 -100,readerWait 变成 0,此时会唤醒在 writerSem 上沉睡的协程 w1。

Q5: 在写锁被拿到的情况下,新协程拿读锁,如何让新协程被阻塞?
func (rw *RWMutex) RLock() {
	if atomic.AddInt32(&rw.readerCount, 1) < 0 {
		// A writer is pending, wait for it.
		runtime_SemacquireMutex(&rw.readerSem, false, 0)
	}
}

继续上面的场景,readerCount = -100 + 1 = -99 < 0。

新的读协程 r1 被沉睡在 readerSem 下面。

假设此时再来一个读协程 r2,则 readerCount = -98,依旧沉睡。

Q6: 在写锁被拿到的情况下,新协程拿读锁,写锁协程释放,如何唤醒等待的读锁协程?

继续上面的场景,此时协程 w1 释放写锁

func (rw *RWMutex) Unlock() {
	// Announce to readers there is no active writer.
	r := atomic.AddInt32(&rw.readerCount, rwmutexMaxReaders)
	// Unblock blocked readers, if any.
	for i := 0; i < int(r); i++ {
		runtime_Semrelease(&rw.readerSem, false, 0)
	}
	// Allow other writers to proceed.
	rw.w.Unlock()
}

在执行

atomic.AddInt32(&rw.readerCount, rwmutexMaxReaders)

后,r = readerCount = -98 + 100 = 2,代表此时有两个读协程 r1 和 r2 在等待

ps: 如果此时有一些新的协程想要拿读锁,他会因为 readerCount = 2 + 1 = 3 > 0 而顺利执行下去,不会被阻塞

之后 for 循环执行两次,将协程 r1 和 协程 r2 都唤醒了。

Q7: 在写锁被拿到的情况下,有两个协程分别去抢读锁和写锁,当写锁被释放时,这两个协程谁会胜利?
func (rw *RWMutex) Unlock() {
	// Announce to readers there is no active writer.
	r := atomic.AddInt32(&rw.readerCount, rwmutexMaxReaders)
	// Unblock blocked readers, if any.
	for i := 0; i < int(r); i++ {
		runtime_Semrelease(&rw.readerSem, false, 0)
	}
	// Allow other writers to proceed.
	rw.w.Unlock()
}

由于是先唤醒读锁,再调用 w.Unlock() ,因此肯定是读协程先胜利!

认为写的比较巧妙的两个点
  1. readerCount 与 rwmutexMaxReaders 的纠缠

    通过 readerCount + rwmutexMaxReaders 以及 readerCount - rwmutexMaxReaders 这两个操作可以得知当前是否有协程等待/持有写锁以及当前等待/持有读锁的协程数量

  2. readerCount 与 readerWait 的纠缠

    在 Lock() 时直接将 readerCount 的值赋给 readerWait,在 readerWait = 0 而非 readerCount = 0 是唤醒写协程,可以避免在 Lock() 后来达到的读协程先于写协程被执行。

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