时间：2022-09-26 09:35:56 | 栏目：Golang | 点击：次
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 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
}
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)
}
}
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()
}
func (rw *RWMutex) RLock() {
if atomic.AddInt32(&rw.readerCount, 1) < 0 {
// A writer is pending, wait for it.
runtime_SemacquireMutex(&rw.readerSem, false, 0)
}
}
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)
}
}
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
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
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。
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。
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。
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。
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！
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
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。