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sync: don't use volatile in Mutex
Volatile loads/stors are only useful for communication with interrupts
or for memory-mapped I/O. They do not provide any sort of safety for
sync.Mutex, while making it *appear* as if it is more safe.
* `sync.Mutex` cannot be used safely inside interrupts, because any
blocking calls (including `Lock`) will cause a runtime panic.
* For multithreading, `volatile` is also the wrong choice. Atomic
operations should be used instead, and the current code would not
work for multithreaded programs anyway.
This commit is contained in:
+6
-23
@@ -3,12 +3,10 @@ package sync
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import (
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"internal/task"
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_ "unsafe"
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"runtime/volatile"
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)
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type Mutex struct {
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state uint8 // Set to non-zero if locked.
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locked bool
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blocked task.Stack
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}
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@@ -16,18 +14,18 @@ type Mutex struct {
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func scheduleTask(*task.Task)
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func (m *Mutex) Lock() {
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if m.islocked() {
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if m.locked {
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// Push self onto stack of blocked tasks, and wait to be resumed.
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m.blocked.Push(task.Current())
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task.Pause()
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return
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}
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m.setlock(true)
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m.locked = true
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}
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func (m *Mutex) Unlock() {
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if !m.islocked() {
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if !m.locked {
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panic("sync: unlock of unlocked Mutex")
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}
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@@ -35,7 +33,7 @@ func (m *Mutex) Unlock() {
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if t := m.blocked.Pop(); t != nil {
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scheduleTask(t)
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} else {
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m.setlock(false)
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m.locked = false
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}
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}
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@@ -45,28 +43,13 @@ func (m *Mutex) Unlock() {
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// and use of TryLock is often a sign of a deeper problem
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// in a particular use of mutexes.
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func (m *Mutex) TryLock() bool {
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if m.islocked() {
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if m.locked {
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return false
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}
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m.Lock()
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return true
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}
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func (m *Mutex) islocked() bool {
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return volatile.LoadUint8(&m.state) != 0
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}
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func (m *Mutex) setlock(b bool) {
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volatile.StoreUint8(&m.state, boolToU8(b))
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}
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func boolToU8(b bool) uint8 {
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if b {
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return 1
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}
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return 0
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}
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type RWMutex struct {
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// waitingWriters are all of the tasks waiting for write locks.
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waitingWriters task.Stack
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