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https://github.com/impact-eintr/netstack.git
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81 lines
2.2 KiB
Go
81 lines
2.2 KiB
Go
package tcp
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import (
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"netstack/tcpip/header"
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"netstack/tcpip/seqnum"
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)
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const (
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// MaxSACKBlocks 是接收端存储的最大SACK数
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MaxSACKBlocks = 6
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)
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// UpdateSACKBlocks 更新SACK块列表以包含 segStart-segEnd 指定的段，只有没有被消费掉的seg才会被用来更新sack。
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// 如果该段恰好是无序传递，那么sack.blocks中的第一个块总是包括由 segStart-segEnd 标识的段。
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func UpdateSACKBlocks(sack *SACKInfo, segStart seqnum.Value, segEnd seqnum.Value, rcvNxt seqnum.Value) {
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newSB := header.SACKBlock{Start: segStart, End: segEnd}
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if sack.NumBlocks == 0 {
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sack.Blocks[0] = newSB
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sack.NumBlocks = 1
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return
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}
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var n = 0
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for i := 0; i < sack.NumBlocks; i++ {
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start, end := sack.Blocks[i].Start, sack.Blocks[i].End
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if end.LessThanEq(start) || start.LessThanEq(rcvNxt) {
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// Discard any invalid blocks where end is before start
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// and discard any sack blocks that are before rcvNxt as
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// those have already been acked.
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continue
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}
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if newSB.Start.LessThanEq(end) && start.LessThanEq(newSB.End) {
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// Merge this SACK block into newSB and discard this SACK
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// block.
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if start.LessThan(newSB.Start) {
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newSB.Start = start
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}
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if newSB.End.LessThan(end) {
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newSB.End = end
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}
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} else {
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// Save this block.
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sack.Blocks[n] = sack.Blocks[i]
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n++
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}
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}
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if rcvNxt.LessThan(newSB.Start) {
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// If this was an out of order segment then make sure that the
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// first SACK block is the one that includes the segment.
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//
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// See the first bullet point in
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// https://tools.ietf.org/html/rfc2018#section-4
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if n == MaxSACKBlocks {
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// If the number of SACK blocks is equal to
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// MaxSACKBlocks then discard the last SACK block.
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n--
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}
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for i := n - 1; i >= 0; i-- {
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sack.Blocks[i+1] = sack.Blocks[i]
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}
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sack.Blocks[0] = newSB
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n++
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}
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sack.NumBlocks = n
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}
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// TrimSACKBlockList 通过删除/修改 start为 < rcvNext 的任何块来更新sack块列表
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func TrimSACKBlockList(sack *SACKInfo, rcvNxt seqnum.Value) {
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n := 0
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for i := 0; i < sack.NumBlocks; i++ { // 遍历
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if sack.Blocks[i].End.LessThanEq(rcvNxt) {
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continue
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}
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if sack.Blocks[i].Start.LessThan(rcvNxt) {
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sack.Blocks[i].Start = rcvNxt
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}
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sack.Blocks[n] = sack.Blocks[i]
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n++
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}
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sack.NumBlocks = n
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}
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