分片机制开始实现 大致是使用了一个map管理+小根堆自动排序实现的

tcpip/network/fragmentation/fragmentation.go

@@ -22,16 +22,18 @@ const HighFragThreshold = 4 << 20 // 4MB
 // net.ipv4.ipfrag_low_thresh for more information.
 const LowFragThreshold = 3 << 20 // 3MB
 
+// Fragmentation 分片处理器对象
 type Fragmentation struct {
 	mu           sync.Mutex
 	highLimit    int
 	lowLimit     int
-	reassemblers map[uint32]*reassembler
+	reassemblers map[uint32]*reassembler // IP报文hash:重组器
 	rList        reassemblerList
 	size         int
 	timeout      time.Duration
 }
 
+// NewFragmentation 新建一个分片处理器
 func NewFragmentation(highMemoryLimit, lowMemoryLimit int, reassemblingTimeout time.Duration) *Fragmentation {
 	if lowMemoryLimit >= highMemoryLimit {
 		lowMemoryLimit = highMemoryLimit
@@ -49,20 +51,41 @@ func NewFragmentation(highMemoryLimit, lowMemoryLimit int, reassemblingTimeout t
 	}
 }
 
+// Process 处理ip报文分片
 func (f *Fragmentation) Process(id uint32, first, last uint16, more bool, vv buffer.VectorisedView) (buffer.VectorisedView, bool) {
-	log.Println("分片机制工作中", id, first, last, vv.First())
 	f.mu.Lock()
 	r, ok := f.reassemblers[id]
-	if ok && r.tooOld(f.timeout) {
+	if ok && r.tooOld(f.timeout) { // 检测一个分片是否存在超过了30s
 		// This is very likely to be an id-collision or someone performing a slow-rate attack.
-		//f.release(r)
+		f.release(r)
 		ok = false
 	}
-	if !ok {
+	if !ok { // 首次注册该报文的分片
 		r = newReassembler(id)
 		f.reassemblers[id] = r
 		f.rList.PushFront(r)
 	}
 	f.mu.Unlock()
+
+	//res, done, consumed := r.process(first, last, more, vv)
+	log.Printf("[%d]的分片 [%d,%d) 合并中\n", id, first, last)
+	r.process(first, last, more, vv)
+
 	return buffer.VectorisedView{}, false
 }
+
+func (f *Fragmentation) release(r *reassembler) {
+	// Before releasing a fragment we need to check if r is already marked as done.
+	// Otherwise, we would delete it twice.
+	if r.checkDoneOrMark() {
+		return
+	}
+
+	delete(f.reassemblers, r.id)
+	f.rList.Remove(r)
+	f.size -= r.size
+	if f.size < 0 {
+		log.Printf("memory counter < 0 (%d), this is an accounting bug that requires investigation", f.size)
+		f.size = 0
+	}
+}