@GraHal, looking at code, early days still, going to take a while to fully understand it to find loose nut.

I inserted this code at line 28, after setup before main IF block.

If error is an undefined element, then pre-defining the array elements could avoid it.

Tried algo normal a few times, keep stopping straight away.

With added code, been running for about 30minutes, no entries taken.

Don’t yet know if pre-setting element has screwed with operation so never reach error or what.

But this could lean towards the  undefined element error option.

if barindex = 0 then
for i = 0 to 100
$Tablevalue2[i] = 0
$val[i] = 0
$val1[i] = 0
next
endif

Running on 1 minute, DAX40(DFB)

Set some demo bots loose with predefined arrays of 100, 10  and 4, all still running from yesterday, however no entries.

That appears to conclude that original error trigger when element[4] was accessed, and undefined.

In THE original code,  ‘CounterBars = 500’, variable indirectly sets the index to 0-488, providing more data points.

This in theory explains why elements undefined, but doesn’t avoid error in tests, had to set A3 =  2000 before bot would stay running.

So another angle to look at.

Could be, value too similar to ignore. been running for an hour or so now, no entries yet on any.

Backtest version, no pre-define, showing entries, so i’m wondering if , pre-defining elements, though avoiding stop error, has affected code in a way, when accessing elements which just hold the default value of zero, and has change the operation. And that may be that back test results could be erroneous for the same/different reason, since not throwing an error.

On a slightly different note, testing, trying to capture an error, and algo has limited/no feedback capabilities, print, graph, etc

In this case, the error pointed to the ‘array’ with message and several possible reasons why.

If you suspect, in this case, that its trying to access a undefined element. Then setting a IF block to check for a valid index range before use, if its a fail, then using another, different command who’s purpose is to fail, like  test = highest[-1], negative index. If IF condition  is out of bounds, then IF code run, which produces a failure before original error, The error message is now different, pointing to something else, so you can use this change, as an indication on what’s maybe happening in the bot when it stops, based on the condition you set.

Not got to a definitive answer but, I think the reason for the error, and what caused it are now clear.

However, the exact path taken to the error is still undefined.

#numbers/var    refer to line numbers in your original code posted.

#Vopt    When variable A3 = 15

#5          CountBars = A3, and then

#27        shift = CountBars-11-1 which equates to 3, giving a shift range of 0,1,2,3.

#29       main IF block is true when barindex is greater than Countbars

#30       main While loop, loops, shift value and, exit condition true when shift smaller than zero

#123     decrementing of shift by -1, end up at shift = -1, which breaks the loop.

Additionally…

#Vopt      A2 = 1.5

#4          Risk = A2

#Vopt     A23 = 77

#Vopt     A24 = 15

#23         X1 = A23 + Risk      which equates to 78.5,     upper level threshold

#24         X2 = A24 – Risk     which equates to 13.5,      lower level threshold

#79       calculates the value of Value2 and then

#80      store Value2 in array, both related to shift value

#85     tests if current value of Value2 is smaller than the lower level threshold of X2, if it is….

#86    the value of i1 is set to one.

#87    The while loop, looks back and test if, pointed to, array value, of Value2, condition falls in between the upper and lower threshold levels.

So, current bar is below X2, #85, lookback searching for last time above X2 #87.

Now here is where I think the error comes in…

i1 is set to 1 #86 , so the look back starts at previous bar to shift , shift+1.

On the first run of main while, shift = 3, and that is the index value used to store arrays, then

adding 1 to shift points to an undefined element ‘4’ and hence error as loop commences.

Now pre-defining arrays to avoid this error means that the default value of these extra elements hold the value of zero.

When checking for the last inside of range, a value of zero, means that may not happen, because all higher elements are 0 and depending on situation at the time.

It’s possible that, if shift = 0, when all this happens, then condition for inside threshold area may be met before i1 increments to a value which results in accessing the undefined element. So not a total failure.

The original code had CountBars = 500, this would have resulted in a shift range of 0 – 488.

In the while loop, looking back would have a lot of elements defined. This avoiding error or at least pushes to an edge case where if, value2 stayed below or above the threshold levels that long and other factors allowed access beyond defined array element could happen.

I suppose the number of consecutive bars, above or below the threshold levels, in that process, is the cut off point.

And depending on timeframe used, the amount of these bars are going to be more or less.

Therefore A3 needs increasing to a value that avoids this array look back error situation.