Don't send initial index twice, use more fetchers

model.go

@@ -42,8 +42,7 @@ var (
 )
 
 const (
-	RemoteFetchers = 4
+	FlagDeleted = 1 << 12
-	FlagDeleted    = 1 << 12
 
 	idxBcastHoldtime = 15 * time.Second  // Wait at least this long after the last index modification
 	idxBcastMaxDelay = 120 * time.Second // Unless we've already waited this long
@@ -51,12 +50,13 @@ const (
 
 func NewModel(dir string) *Model {
 	m := &Model{
-		dir:    dir,
+		dir:          dir,
-		global: make(map[string]File),
+		global:       make(map[string]File),
-		local:  make(map[string]File),
+		local:        make(map[string]File),
-		remote: make(map[string]map[string]File),
+		remote:       make(map[string]map[string]File),
-		need:   make(map[string]bool),
+		need:         make(map[string]bool),
-		nodes:  make(map[string]*protocol.Connection),
+		nodes:        make(map[string]*protocol.Connection),
+		lastIdxBcast: time.Now(),
 	}
 
 	go m.printStats()

model_puller.go

@@ -11,10 +11,6 @@ held for as short a time as possible.
 
 TODO(jb): Refactor this into smaller and cleaner pieces.
 
-TODO(jb): Some kind of coalescing / rate limiting of index sending, so we don't
-send hundreds of index updates in a short period if time when deleting files
-etc.
-
 */
 
 import (
@@ -29,6 +25,8 @@ import (
 	"github.com/calmh/syncthing/buffers"
 )
 
+const RemoteFetchers = 8
+
 func (m *Model) pullFile(name string) error {
 	m.RLock()
 	var localFile = m.local[name]
@@ -61,7 +59,7 @@ func (m *Model) pullFile(name string) error {
 	local, remote := localFile.Blocks.To(globalFile.Blocks)
 	var fetchDone sync.WaitGroup
 
-	// One local copy routing
+	// One local copy routine
 
 	fetchDone.Add(1)
 	go func() {