Exit and Enter right away

[entropy]

I'm really stumped on this issue and am wondering if exiting a position also causes a "return;" to happen and kill a process.

Upon signal==0, I want this code to flip any open short position to a long position if lettradefinish==false (and cancel the signal if lettradefinish==true). if lettradefinish==true it cancels the signal correctly. But if lettradefinish==false, it just exits the short position while not opening the new long one.

assumptions:
Lettradefinish is an arbitrarily set bool that i have permanently set to false
signal==0 means the rest of my code generated a signal to go long (default for no signal is signal==-1), assume signal==0
AllowTrade is true
we head into this code with a short position

if (signal==0)
{
if (Position.MarketPosition==MarketPosition.Short)
{
if(!LetTradeFinish)
{
Position.Close();

}
else
{
signal=-1;
}
}

if(AllowTrade && (Position.MarketPosition==MarketPosition.Flat || (Position.Quantity<Quantity*this.EntriesPerDirecti on && AllowMultiEntries)))
{
IOrder order = this.EnterLong(0,Quantity,"Long Entry");
}
}

Would greatly appreciate any help as I think I must not understand what some of these order lines are doing from a process/order of events perspective.

[NinjaTrader_Austin]

entropy, please simplify your code until it is function correctly and then you can start layering on the complexity again. I don't have you full code, so I can't run it, but I have a feeling all of the long order entry conditions aren't being met.

[entropy]

Hi Austin, your comments helped me focus on the values and I've solved the problem.

I was thinking that as soon as I first exited the existing short position that marketposition would immediately be marketposition.flat, but this must not be the case until the next tick.

Instead of counting on marketposition.flat, i'm using a bool to track if the exit happened yet and to allow the long entry if it did, and it works great.

Thanks!