pytest: Run sendpay in a future so it doesn't complain about failing

For the permfail tests the sendpay call is supposed to fail, so this
was printing stacktraces upon success. Running in futures captures any
thrown exceptions and rethrows them when calling `result()`, in our
case we just ignore them.

Signed-off-by: Christian Decker <decker.christian@gmail.com>

tests/test_lightningd.py

@@ -199,14 +199,12 @@ class LightningDTests(BaseLightningDTests):
         rhash = ldst.rpc.invoice(amt, label)['rhash']
         assert ldst.rpc.listinvoice(label)[0]['complete'] == False
 
-        def call_pay():
-            routestep = {
-                'msatoshi' : amt,
-                'id' : ldst.info['id'],
-                'delay' : 5,
-                'channel': '1:1:1'
-            }
-            lsrc.rpc.sendpay(to_json([routestep]), rhash, async=False)
+        routestep = {
+            'msatoshi' : amt,
+            'id' : ldst.info['id'],
+            'delay' : 5,
+            'channel': '1:1:1'
+        }
 
         def wait_pay():
             # Up to 10 seconds for payment to succeed.
@@ -217,12 +215,10 @@ class LightningDTests(BaseLightningDTests):
                 time.sleep(0.1)
 
         if async:
-            t = threading.Thread(target=call_pay)
-            t.daemon = True
-            t.start()
+            self.executor.submit(lsrc.rpc.sendpay, to_json([routestep]), rhash, async=False)
             return self.executor.submit(wait_pay)
         else:
-            call_pay()
+            lsrc.rpc.sendpay(to_json([routestep]), rhash, async=False)
 
     def test_connect(self):
         l1,l2 = self.connect()
@@ -469,7 +465,7 @@ class LightningDTests(BaseLightningDTests):
         self.fund_channel(l1, l2, 10**6)
 
         # This will fail at l2's end.
-        t=self.pay(l1,l2,200000000,async=True)
+        t=self.pay(l1, l2, 200000000, async=True)
 
         l2.daemon.wait_for_log('dev_disconnect permfail')
         l2.daemon.wait_for_log('sendrawtx exit 0')