lnwallet: export HTLC generating methods

This commit is contained in:
Johan T. Halseth committed 2021-03-05 10:58:42 +01:00
1 parent 3a3076397a
commit 2bc37db61a
2 files changed
+10 -10

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@@ -35,7 +35,7 @@ var (
TimelockShift = uint32(1 << 29)
)
// createHtlcSuccessTx creates a transaction that spends the output on the
// CreateHtlcSuccessTx creates a transaction that spends the output on the
// commitment transaction of the peer that receives an HTLC. This transaction
// essentially acts as an off-chain covenant as it's only permitted to spend
// the designated HTLC output, and also that spend can _only_ be used as a
@@ -45,7 +45,7 @@ var (
// In order to spend the HTLC output, the witness for the passed transaction
// should be:
// * <0> <sender sig> <recvr sig> <preimage>
func createHtlcSuccessTx(chanType channeldb.ChannelType,
func CreateHtlcSuccessTx(chanType channeldb.ChannelType,
htlcOutput wire.OutPoint, htlcAmt btcutil.Amount, csvDelay uint32,
revocationKey, delayKey *btcec.PublicKey) (*wire.MsgTx, error) {
@@ -85,7 +85,7 @@ func createHtlcSuccessTx(chanType channeldb.ChannelType,
return successTx, nil
}
// createHtlcTimeoutTx creates a transaction that spends the HTLC output on the
// CreateHtlcTimeoutTx creates a transaction that spends the HTLC output on the
// commitment transaction of the peer that created an HTLC (the sender). This
// transaction essentially acts as an off-chain covenant as it spends a 2-of-2
// multi-sig output. This output requires a signature from both the sender and
@@ -101,7 +101,7 @@ func createHtlcSuccessTx(chanType channeldb.ChannelType,
// NOTE: The passed amount for the HTLC should take into account the required
// fee rate at the time the HTLC was created. The fee should be able to
// entirely pay for this (tiny: 1-in 1-out) transaction.
func createHtlcTimeoutTx(chanType channeldb.ChannelType,
func CreateHtlcTimeoutTx(chanType channeldb.ChannelType,
htlcOutput wire.OutPoint, htlcAmt btcutil.Amount,
cltvExpiry, csvDelay uint32,
revocationKey, delayKey *btcec.PublicKey) (*wire.MsgTx, error) {