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authorSebastian Falbesoner <sebastian.falbesoner@gmail.com>2021-09-16 14:32:43 +0200
committerSebastian Falbesoner <sebastian.falbesoner@gmail.com>2021-09-16 14:32:43 +0200
commit0f275246027266fa256d0a09251bb2c88d9bd72f (patch)
tree0c0bd12fe975e1f76029a873f725ce479bdb9013 /test
parentd1e2481274edf2ac14747be633d86ecd46814ef4 (diff)
downloadbitcoin-0f275246027266fa256d0a09251bb2c88d9bd72f.tar.xz
test: scale amounts in test_doublespend_tree down by factor 10
This is done in order to prepare the make_utxo helper to use MiniWallet, which only supports creating transactions with single inputs, i.e. we need to create amounts small enough to be funded by coinbase transactions (50 BTC).
Diffstat (limited to 'test')
-rwxr-xr-xtest/functional/feature_rbf.py12
1 files changed, 6 insertions, 6 deletions
diff --git a/test/functional/feature_rbf.py b/test/functional/feature_rbf.py
index d4e483ccd5..c9eb5a8919 100755
--- a/test/functional/feature_rbf.py
+++ b/test/functional/feature_rbf.py
@@ -201,10 +201,10 @@ class ReplaceByFeeTest(BitcoinTestFramework):
def test_doublespend_tree(self):
"""Doublespend of a big tree of transactions"""
- initial_nValue = 50 * COIN
+ initial_nValue = 5 * COIN
tx0_outpoint = self.make_utxo(self.nodes[0], initial_nValue)
- def branch(prevout, initial_value, max_txs, tree_width=5, fee=0.0001 * COIN, _total_txs=None):
+ def branch(prevout, initial_value, max_txs, tree_width=5, fee=0.00001 * COIN, _total_txs=None):
if _total_txs is None:
_total_txs = [0]
if _total_txs[0] >= max_txs:
@@ -235,7 +235,7 @@ class ReplaceByFeeTest(BitcoinTestFramework):
_total_txs=_total_txs):
yield x
- fee = int(0.0001 * COIN)
+ fee = int(0.00001 * COIN)
n = MAX_REPLACEMENT_LIMIT
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n)
@@ -248,10 +248,10 @@ class ReplaceByFeeTest(BitcoinTestFramework):
# This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, dbl_tx_hex, 0)
- # 1 BTC fee is enough
+ # 0.1 BTC fee is enough
dbl_tx = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)]
- dbl_tx.vout = [CTxOut(initial_nValue - fee * n - 1 * COIN, DUMMY_P2WPKH_SCRIPT)]
+ dbl_tx.vout = [CTxOut(initial_nValue - fee * n - int(0.1 * COIN), DUMMY_P2WPKH_SCRIPT)]
dbl_tx_hex = dbl_tx.serialize().hex()
self.nodes[0].sendrawtransaction(dbl_tx_hex, 0)
@@ -264,7 +264,7 @@ class ReplaceByFeeTest(BitcoinTestFramework):
# Try again, but with more total transactions than the "max txs
# double-spent at once" anti-DoS limit.
for n in (MAX_REPLACEMENT_LIMIT + 1, MAX_REPLACEMENT_LIMIT * 2):
- fee = int(0.0001 * COIN)
+ fee = int(0.00001 * COIN)
tx0_outpoint = self.make_utxo(self.nodes[0], initial_nValue)
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n)