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test: scale amounts in test_doublespend_tree down by factor 10

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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).
This commit is contained in:
Sebastian Falbesoner 2021-09-16 14:32:43 +02:00
parent d1e2481274
commit 0f27524602
1 changed files with 6 additions and 6 deletions
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12
test/functional/feature_rbf.py
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@ -201,10 +201,10 @@ class ReplaceByFeeTest(BitcoinTestFramework):
def test_doublespend_tree(self): def test_doublespend_tree(self):
"""Doublespend of a big tree of transactions""" """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) 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: if _total_txs is None:
_total_txs = [0] _total_txs = [0]
if _total_txs[0] >= max_txs: if _total_txs[0] >= max_txs:
@ -235,7 +235,7 @@ class ReplaceByFeeTest(BitcoinTestFramework):
_total_txs=_total_txs): _total_txs=_total_txs):
yield x yield x
fee = int(0.0001 * COIN) fee = int(0.00001 * COIN)
n = MAX_REPLACEMENT_LIMIT n = MAX_REPLACEMENT_LIMIT
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee)) tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n) assert_equal(len(tree_txs), n)
@ -248,10 +248,10 @@ class ReplaceByFeeTest(BitcoinTestFramework):
# This will raise an exception due to insufficient fee # This will raise an exception due to insufficient fee
assert_raises_rpc_error(-26, "insufficient fee", self.nodes[0].sendrawtransaction, dbl_tx_hex, 0) 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 = CTransaction()
dbl_tx.vin = [CTxIn(tx0_outpoint, nSequence=0)] 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() dbl_tx_hex = dbl_tx.serialize().hex()
self.nodes[0].sendrawtransaction(dbl_tx_hex, 0) 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 # Try again, but with more total transactions than the "max txs
# double-spent at once" anti-DoS limit. # double-spent at once" anti-DoS limit.
for n in (MAX_REPLACEMENT_LIMIT + 1, MAX_REPLACEMENT_LIMIT * 2): 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) tx0_outpoint = self.make_utxo(self.nodes[0], initial_nValue)
tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee)) tree_txs = list(branch(tx0_outpoint, initial_nValue, n, fee=fee))
assert_equal(len(tree_txs), n) assert_equal(len(tree_txs), n)