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7c3991ee2f
* Use a FIFO queue instead of a channel to reduce backpressure * Make sure someone wakes up * Tweaks * Add comments
65 lines
1.3 KiB
Go
65 lines
1.3 KiB
Go
package input
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import (
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"sync"
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)
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type fifoQueue struct {
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tasks []*inputTask
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count int
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mutex sync.Mutex
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notifs chan struct{}
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}
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func newFIFOQueue() *fifoQueue {
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q := &fifoQueue{
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notifs: make(chan struct{}, 1),
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}
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return q
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}
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func (q *fifoQueue) push(frame *inputTask) {
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q.mutex.Lock()
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defer q.mutex.Unlock()
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q.tasks = append(q.tasks, frame)
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q.count++
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select {
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case q.notifs <- struct{}{}:
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default:
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}
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}
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// pop returns the first item of the queue, if there is one.
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// The second return value will indicate if a task was returned.
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// You must check this value, even after calling wait().
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func (q *fifoQueue) pop() (*inputTask, bool) {
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q.mutex.Lock()
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defer q.mutex.Unlock()
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if q.count == 0 {
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return nil, false
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}
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frame := q.tasks[0]
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q.tasks[0] = nil
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q.tasks = q.tasks[1:]
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q.count--
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if q.count == 0 {
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// Force a GC of the underlying array, since it might have
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// grown significantly if the queue was hammered for some reason
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q.tasks = nil
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}
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return frame, true
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}
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// wait returns a channel which can be used to detect when an
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// item is waiting in the queue.
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func (q *fifoQueue) wait() <-chan struct{} {
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q.mutex.Lock()
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defer q.mutex.Unlock()
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if q.count > 0 && len(q.notifs) == 0 {
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ch := make(chan struct{})
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close(ch)
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return ch
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}
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return q.notifs
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}
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