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it seems to make sense since, in each place, the Key and Priority represent the same information b/c you know the saying... "It is better to have 100 functions operate on one data structure than 10 functions on 10 data structures." License: MIT Signed-off-by: Brian Tiger Chow <brian@perfmode.com>
79 lines
1.9 KiB
Go
79 lines
1.9 KiB
Go
package strategy
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import (
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wantlist "github.com/jbenet/go-ipfs/exchange/bitswap/wantlist"
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peer "github.com/jbenet/go-ipfs/peer"
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u "github.com/jbenet/go-ipfs/util"
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)
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// TODO: at some point, the strategy needs to plug in here
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// to help decide how to sort tasks (on add) and how to select
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// tasks (on getnext). For now, we are assuming a dumb/nice strategy.
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type taskQueue struct {
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tasks []*task
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taskmap map[string]*task
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}
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func newTaskQueue() *taskQueue {
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return &taskQueue{
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taskmap: make(map[string]*task),
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}
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}
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type task struct {
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Entry wantlist.Entry
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Target peer.Peer
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}
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// Push currently adds a new task to the end of the list
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// TODO: make this into a priority queue
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func (tl *taskQueue) Push(block u.Key, priority int, to peer.Peer) {
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if task, ok := tl.taskmap[taskKey(to, block)]; ok {
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// TODO: when priority queue is implemented,
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// rearrange this task
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task.Entry.Priority = priority
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return
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}
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task := &task{
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Entry: wantlist.Entry{
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Key: block,
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Priority: priority,
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},
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Target: to,
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}
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tl.tasks = append(tl.tasks, task)
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tl.taskmap[taskKey(to, block)] = task
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}
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// Pop 'pops' the next task to be performed. Returns nil no task exists.
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func (tl *taskQueue) Pop() *task {
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var out *task
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for len(tl.tasks) > 0 {
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// TODO: instead of zero, use exponential distribution
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// it will help reduce the chance of receiving
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// the same block from multiple peers
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out = tl.tasks[0]
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tl.tasks = tl.tasks[1:]
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delete(tl.taskmap, taskKey(out.Target, out.Entry.Key))
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// Filter out blocks that have been cancelled
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if out.Entry.Priority >= 0 { // FIXME separate the "cancel" signal from priority
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break
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}
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}
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return out
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}
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// Remove lazily removes a task from the queue
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func (tl *taskQueue) Remove(k u.Key, p peer.Peer) {
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t, ok := tl.taskmap[taskKey(p, k)]
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if ok {
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t.Entry.Priority = -1
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}
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}
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// taskKey returns a key that uniquely identifies a task.
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func taskKey(p peer.Peer, k u.Key) string {
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return string(p.Key() + k)
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}
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