scooby-doo BOJ JPY pairs trading robot

Quoting SteveHopwood

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ptarmigan, many many apologies for not replying. I managed to miss both of your postings of this; I saw it a few seconds ago because I haven't the faintest idea what update to the indi scoobs is talking about and was looking back through the posts to try to find out.

ptarmigan is suggesting a call to FFCal every 5 minutes instead of every minute. Sounds like a plan to me. Any one object?

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Quoting SteveHopwood

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Here we go. Would the coders here like to check the logic so we don't waste time tonight? Sing out if you see anything wrong.



[code]...

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Quoting scooby-doo

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Steve,

You going to update the indicator aswell ?

Scoobs

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Quoting scooby-doo

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Steve,

Is that break; correct ? or are we going to get stuck in the while() loop ?

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   if (OldBars != Bars)
   {
      Comment(Gap + "New candle. Waiting for all new candles to form");
      bool rollover = false;
      while (!rollover)
      {
         rollover = true;
         for (int cc = 0; cc < NoOfPairs; cc++)
         {
            if (LastTime[cc] == iTime(TradePair[cc], 0, 0) && MarketInfo(TradePair[cc], MODE_SPREAD) > 0 
                                      && MarketInfo(TradePair[cc], MODE_SWAPLONG) != 0)
            {
               rollover = false;
               break;
            }//if (lastTime[cc] == iTime(symbols[cc], 0, 0)
         }//for (int cc = 0; cc < NoOfPairs; cc++)         
      }//while (!rollover)
   
      // now all candles are new and in lockstep so update stored timestamps      
      for (cc = 0; cc < NoOfPairs; cc++) 
      {   
         LastTime[cc] = iTime(TradePair[cc], 0, 0);
      }//for (cc = 0; cc < NoOfPairs; cc++) 
      
      OldBars = Bars;
   }//if (OldBars != Bars)

Quoting SteveHopwood

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I think the break is ok as it aborts the for:next loop and returns contro to the while() loop.

I was already considering the possibility of an endless loop if the user has a pair entered that the crim does not allow. Here is the updated code.



[code]...

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Quoting scooby-doo

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I dont know but it looks to me as though we only ever check the first pair in the array. ??

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Quoting SteveHopwood

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The rollover = false/break stuff only happens if the conditional is true. If it is, the code only needs to set rollover to false once; there is no need to cycle through the remaining pairs.

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Quoting scooby-doo

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ok, i will take your word for it.

Maybe I am still drunk from lastnight or this morning, god, i dont even know what time it is let alone GMT.

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Quoting SteveHopwood

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Hehe. Have a 'smug' pic ready for when it gets stuck in an endless loop tonight.

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$NoOfPairs = array("USDJPY", "GBPJPY", "EURJPY", "CADJPY", "CHFJPY", "AUDJPY", "NZDJPY");
$rollover = false;
$iTime = array(3, 10, 8, 9, 5, 7, 2);

while(!$rollover)
{
    $rollover = true;
    for($cc=0; $cc < count($NoOfPairs); $cc++)
    {
        if($iTime[$cc] != 10) // ($LastTime[$cc] == $iTime[$cc])
        {
            $rollover = false;
            $iTime[$cc] += 1;
            print("$NoOfPairs[$cc] is not ready.\n");
            break;
        }
        print("$NoOfPairs[$cc] is OK.\n");
    }
} 

Quoting SteveHopwood

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Hehe. Have a 'smug' pic ready for when it gets stuck in an endless loop tonight.

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bool IsNewCandleSet()
{
   int i;
   bool rollover = true;

   for (i = 0; i < 7; i++) 
      if (lastTime[i] == iTime(symbols[i], 0, 0)
         rollover = false;
         
   // early out because no candles have changed
   if (rollover == true)
      return (false);
         
   // at least one candle has changed so now wait for the rest to 
   // rollover
   while (!rollover) {
      rollover = true;
      
      for (i = 0; i < 7; i++) {
         if (lastTime[i] == iTime(symbols[i], 0, 0) {
            rollover = false;
            break;
         }
      }
   }
   
   // now all candles are new and in lockstep so update
   // stored timestamps
   for (i = 0; i < 7; i++) 
      lastTime[i] = iTime(symbols[i], 0, 0);

   return(true);
}

Quoting Pippopotamus

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If you please, 200 SMA as snap line on which time frame ?

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Quoting Pippopotamus

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Thanks Scooby, and thanks for all you have done here.

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