2 ticks offset strategy
Hi,
I don't know C++ and need some help in implementing one strategy.
I want to enter long 2 ticks below the ask generated by one of the Kwikpop indicators. This indicator = 1, only for the signal bar when it was generated.
Of course, 2 ticks below the ask could be generated on the current bar, or on the next one, or the following one or never.
So, it looks like I have to save the ask on the signal bar and then subtract 0.5. I'm not sure I'm doing it correctly.
That's the excerpt:
/// <summary>
/// Called on each bar update event (incoming tick)
/// </summary>
protected override void OnBarUpdate()
{
double x;
///condition set 1 (trying to get the current
///ask and save it but how?!)
if (N [0] == 1)
{
x = GetCurrentAsk();
}
/// Condition set 2
/// if the condition of 2 ticks below comes, let's say, only on the second
/// bar after signal, then N[0] wouldn't be equal 1, but
///on the previous bar N[1] was equal 1.
if (N[0] == 1
||
N[1] == 1
||
N[2] == 1)
{
EnterLongLimit(2,double.x-0.5);
}
Unfortuanately, there are compilations errors.
I'd appreciate your reply.
Alex. [email protected]
I don't know C++ and need some help in implementing one strategy.
I want to enter long 2 ticks below the ask generated by one of the Kwikpop indicators. This indicator = 1, only for the signal bar when it was generated.
Of course, 2 ticks below the ask could be generated on the current bar, or on the next one, or the following one or never.
So, it looks like I have to save the ask on the signal bar and then subtract 0.5. I'm not sure I'm doing it correctly.
That's the excerpt:
/// <summary>
/// Called on each bar update event (incoming tick)
/// </summary>
protected override void OnBarUpdate()
{
double x;
///condition set 1 (trying to get the current
///ask and save it but how?!)
if (N [0] == 1)
{
x = GetCurrentAsk();
}
/// Condition set 2
/// if the condition of 2 ticks below comes, let's say, only on the second
/// bar after signal, then N[0] wouldn't be equal 1, but
///on the previous bar N[1] was equal 1.
if (N[0] == 1
||
N[1] == 1
||
N[2] == 1)
{
EnterLongLimit(2,double.x-0.5);
}
Unfortuanately, there are compilations errors.
I'd appreciate your reply.
Alex. [email protected]
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