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never use a 6 byte packet number

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According to the IETF draft, the only packet number lengths are 1, 2 and
4 bytes. With the given formula for the packet number derivation, we
would have only sent a 6 byte packet number if the difference between
the largest unacked and the current packet number exceeded 2^31, so this
would never have happened anyway.
This commit is contained in:
Marten Seemann 2017-10-22 00:15:09 +07:00
parent bbe9affa27
commit 6c394acde7
3 changed files with 13 additions and 19 deletions
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12
internal/protocol/packet_number.go
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@ -27,18 +27,14 @@ func delta(a, b PacketNumber) PacketNumber {
return a - b
}
// GetPacketNumberLengthForPublicHeader gets the length of the packet number for the public header
// GetPacketNumberLengthForHeader gets the length of the packet number for the public header
// it never chooses a PacketNumberLen of 1 byte, since this is too short under certain circumstances
func GetPacketNumberLengthForPublicHeader(packetNumber PacketNumber, leastUnacked PacketNumber) PacketNumberLen {
func GetPacketNumberLengthForHeader(packetNumber PacketNumber, leastUnacked PacketNumber) PacketNumberLen {
diff := uint64(packetNumber - leastUnacked)
if diff < (2 << (uint8(PacketNumberLen2)*8 - 2)) {
if diff < (1 << (uint8(PacketNumberLen2)*8 - 1)) {
return PacketNumberLen2
}
if diff < (2 << (uint8(PacketNumberLen4)*8 - 2)) {
return PacketNumberLen4
}
// we do not check if there are less than 2^46 packets in flight, since flow control and congestion control will limit this number *a lot* sooner
return PacketNumberLen6
return PacketNumberLen4
}
// GetPacketNumberLength gets the minimum length needed to fully represent the packet number