diff --git a/test/functional/feature_block.py b/test/functional/feature_block.py index f27d3cb0aa..4fb710e552 100755 --- a/test/functional/feature_block.py +++ b/test/functional/feature_block.py @@ -597,6 +597,8 @@ class FullBlockTest(BitcoinTestFramework): b44.nTime = self.tip.nTime + 1 b44.hashPrevBlock = self.tip.sha256 b44.vtx.append(coinbase) + tx = self.create_and_sign_transaction(out[14], 1) + b44.vtx.append(tx) b44.block_height = height b44.hashMerkleRoot = b44.calc_merkle_root() b44.solve() @@ -687,7 +689,7 @@ class FullBlockTest(BitcoinTestFramework): # Test block timestamps # -> b31 (8) -> b33 (9) -> b35 (10) -> b39 (11) -> b42 (12) -> b43 (13) -> b53 (14) -> b55 (15) # \-> b54 (15) - # + # -> b44 (14)\-> b48 () self.move_tip(43) b53 = self.next_block(53, spend=out[14]) self.send_blocks([b53], False) @@ -707,6 +709,21 @@ class FullBlockTest(BitcoinTestFramework): self.send_blocks([b55], True) self.save_spendable_output() + # The block which was previously rejected because of being "too far(3 hours)" must be accepted 2 hours later. + # The new block is only 1 hour into future now and we must reorg onto to the new longer chain. + # The new bestblock b48p is invalidated manually. + # -> b31 (8) -> b33 (9) -> b35 (10) -> b39 (11) -> b42 (12) -> b43 (13) -> b53 (14) -> b55 (15) + # \-> b54 (15) + # -> b44 (14)\-> b48 () -> b48p () + self.log.info("Accept a previously rejected future block at a later time") + node.setmocktime(int(time.time()) + 2*60*60) + self.move_tip(48) + self.block_heights[b48.sha256] = self.block_heights[b44.sha256] + 1 # b48 is a parent of b44 + b48p = self.next_block("48p") + self.send_blocks([b48, b48p], success=True) # Reorg to the longer chain + node.invalidateblock(b48p.hash) # mark b48p as invalid + node.setmocktime(0) + # Test Merkle tree malleability # # -> b42 (12) -> b43 (13) -> b53 (14) -> b55 (15) -> b57p2 (16) @@ -1345,7 +1362,7 @@ class FullBlockTest(BitcoinTestFramework): tx.calc_sha256() # sign a transaction, using the key we know about - # this signs input 0 in tx, which is assumed to be spending output n in spend_tx + # this signs input 0 in tx, which is assumed to be spending output 0 in spend_tx def sign_tx(self, tx, spend_tx): scriptPubKey = bytearray(spend_tx.vout[0].scriptPubKey) if (scriptPubKey[0] == OP_TRUE): # an anyone-can-spend