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Then draw a horizontal line.
Why do you need any buffers and their recalculation? ObjectSetDoubke(0,myHLine,OBJPROP_PRICE,concrete_price_value);
the user will see the line, you won't waste time recalculating the buffers, everyone is happy and happily disperse
PS/understand correctly - indicator buffer, it's for transferring(sharing/sharing) the results of calculations performed in the first place. Drawing a horizontal line of length N is different
the line should be a curve, i.e. its first value should show current Bid+size, the previous value - Bid+size tick back
And I will need this line to take indicator values for my EAWhat is the correct syntax for a for statement when an expression consists of several parts?
I'm writing it like this:
It doesn't compile.
I write it like this: for(int i=candle_two_index-1,int k=0;i>=candle_two_index-candles_oneside;i--,k++)
Try to compile int i=candle_two_index-1,int k=0; or int i=1,int k=0; separately without loop - you will understand the error
What is the correct syntax for a for statement when an expression consists of several parts?
I'm writing it like this:
It doesn't compile.
One int is redundant:
Good afternoon!
Do not pass by)
Could someone please help me in the following task: add alert to the indicator to trigger after the appearance of the signal on the screen, I would be thousandths of a thank you!
I really tried to do it myself, but it doesn't work.
Good afternoon!
Question on mcl4 syntax: if I set a flag like this:
then the if(!rost) condition will catch only the condition for rost=false or all conditions not falling under rost=true?
Thanks.
Good afternoon!
Question about mcl4 syntax: if I set a flag like this:
then theif(!rost) condition will catch only the condition for rost=false or all conditions not falling under rost=true?
Thanks.
The condition is unambiguous: if rost is false
The condition is unambiguous: if rost is false
What happens to conditions that are neithertrue nor false? Will they be ignored, or for example the conditionif(op1>cl2 && cl1<cl2) is captured? According to Russian logic this is also "NOT". Unfortunately, the handbook does not explicitly state that (!a) = (a=false).
What happens to conditions that are neithertrue nor false? Will they be ignored, or for example the conditionif(op1>cl2 && cl1<cl2) is captured? According to Russian logic this is also "NOT". Unfortunately, the handbook does not explicitly state that (!a) = (a=false).