Shift of the maximum value over a specific number of bars

[Stanfillirenfro]

Hello!
I am sorry to bring a Metatrader’s term to this forum. I am struggling since one week already with this topic, but I am failing to solve the problem. I hope you could help me understanding this term.
I have this function from Metatrader4:

for (int i = 0; i <= 10; i++)
{
double a = iHighest(NULL, 0, MODE_LOW, Period, i);
}

I instantly thought this was equal to higher low in Ninjatrader8, because of iHighest and MODE_LOW. I then set the Period to 1, considering a single bar. “a” should be the low of the corresponding bar. But if I have

double b = iHighest(NULL, 0, MODE_HIGH, Period, i);

, according to my understanding, “b” should be the higher high. Considering a single bar, this should be the high of the bar. So that I have used the function MAX(High, 1) - MAX(Low, 1). With the operation b - a, the result will always be positive, regardless if the bar is bullish or bearish. This is not what I want to have.
In Google, I found the definition of iHighest as followed: Returns the shift of the maximum value over a specific number of bars depending on type.
But I want there my specific number of bars to be 1.
In Ninjatrader documentation, I also found this:
int highestBarsAgo = HighestBar(High, Bars.BarsSinceNewTradingDay);
double highestPrice = High[highestBarsAgo];
With this, I would already be out of range in my loop.
Could someone advise me what could the the equivalent of double a = iHighest(NULL, 0, MODE_LOW, Period, i) in Ninjatrader?
Any help would be very appreciate.

Thanks!

[NinjaTrader_ChrisL]

Hi Stan, thanks for posting.

To find the highest high price over a certain period, use the MAX method. This example gives the highest high price over 20 bars:

double value = MAX(High, 20)[0];

The reason you would use this kind of method is to evaluate what the highest high price is over x amount of bars. If you only need to evaluate one bar, the prices can be accessed in the OHLC price arrays. The lowest period you can pass to MAX is 1 e.g. MAX(High, 1)[0] will return either the last bar or the most recent bar high price. This can also be written Math.Max(High[1], High[0]);

Kind regards,
-ChrisL

[Stanfillirenfro]

Many thanks ChrisL!