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Merge bitcoin/bitcoin#25820: [test] make tx6 child of tx5, not tx3, in rbf_tests

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49db42cdf5 [test] make tx6 child of tx5, not tx3, in rbf_tests (glozow)

Pull request description:

  A small overlooked oopsie from #25674.
  There is no effect on the test results because tx3 and tx5 pay the same fee, but this was the intended configuration, as the comment suggests.

ACKs for top commit:
  instagibbs:
    ACK 49db42cdf5
  darosior:
    Github diff ACK 49db42cdf5. Should have catched this. :/

Tree-SHA512: 2f54337ac3edc38707115cde5b466a85b8a6ac0a0a507effa0e9fecb12c9be196ecd1b16702bc23ba617cfb6a3b5db27d3b71616b3c2dadb186c699c4609831e
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fanquake 2022-08-11 14:19:36 +01:00
commit dd62721ba9
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src/test/rbf_tests.cpp
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@ -76,7 +76,7 @@ BOOST_FIXTURE_TEST_CASE(rbf_helper_functions, TestChain100Setup)
// Create a parent tx5 and child tx6 where both have very low fees
const auto tx5 = make_tx(/*inputs=*/ {m_coinbase_txns[2]}, /*output_values=*/ {1099 * CENT});
pool.addUnchecked(entry.Fee(low_fee).FromTx(tx5));
const auto tx6 = make_tx(/*inputs=*/ {tx3}, /*output_values=*/ {1098 * CENT});
const auto tx6 = make_tx(/*inputs=*/ {tx5}, /*output_values=*/ {1098 * CENT});
pool.addUnchecked(entry.Fee(low_fee).FromTx(tx6));
// Make tx6's modified fee much higher than its base fee. This should cause it to pass
// the fee-related checks despite being low-feerate.