pytest: fix flake in test_htlc_in_timeout with anchors.

If we try to reuse the same UTXO for the HTLC as the anchor, it will clash.
One block later we can spend the anchor change, all good.

Signed-off-by: Rusty Russell <[email protected]>

tests/test_misc.py

@@ -359,7 +359,14 @@ def test_htlc_out_timeout(node_factory, bitcoind, executor):
     # We hit deadline (we give 1 block grace), then mined another.
     assert blocks2 == -2
 
-    bitcoind.generate_block(1, wait_for_mempool=txid)
+    # If we try to reuse the same output as we used for the anchor spend, then
+    # bitcoind can reject it.  In that case we'll try again after we get change
+    # from anchor spend.
+    if txid not in bitcoind.rpc.getrawmempool():
+        bitcoind.generate_block(1)
+        bitcoind.generate_block(1, wait_for_mempool=1)
+    else:
+        bitcoind.generate_block(1)
 
     rawtx, txid, blocks = l1.wait_for_onchaind_tx('OUR_DELAYED_RETURN_TO_WALLET',
                                                   'OUR_HTLC_TIMEOUT_TX/DELAYED_OUTPUT_TO_US')
@@ -424,7 +431,16 @@ def test_htlc_in_timeout(node_factory, bitcoind, executor):
     _, txid, blocks = l2.wait_for_onchaind_tx('OUR_HTLC_SUCCESS_TX',
                                               'OUR_UNILATERAL/THEIR_HTLC')
     assert blocks == 0
-    bitcoind.generate_block(1, wait_for_mempool=txid)
+
+    # If we try to reuse the same output as we used for the anchor spend, then
+    # bitcoind can reject it.  In that case we'll try again after we get change
+    # from anchor spend.
+    if txid not in bitcoind.rpc.getrawmempool():
+        bitcoind.generate_block(1)
+        bitcoind.generate_block(1, wait_for_mempool=1)
+    else:
+        bitcoind.generate_block(1)
+
     rawtx, txid, blocks = l2.wait_for_onchaind_tx('OUR_DELAYED_RETURN_TO_WALLET',
                                                   'OUR_HTLC_SUCCESS_TX/DELAYED_OUTPUT_TO_US')
     assert blocks == 4