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void OpenMarketOrder(int ticket_,string symbol_,int type_,double price_,double lot_)
If yes why using that?
Ok so know that if you call the Print() function too fast repeatedly, especially in fast loops, it can start to skip lines in the log.
I am not sure if that is the problem here but i see a while loop with a print() call in your code.
I do not know how many cycles you loop has but you can for example try the following:
{
Print(IntegerToString(i));
}
When i run this code the log starts at cycle 749
Now if i add a slight delay like this
{
Print(IntegerToString(i));
Sleep(10);
}
Then it does go all the way.
I am not saying that this is the problem but it could be and so it's easy for you to try it by adding a slight delay to your code.
Ok so know that if you call the Print() function too fast repeatedly, especially in fast loops, it can start to skip lines in the log.
I am not sure if that is the problem here but i see a while loop with a print() call in your code.
I do not know how many cycles you loop has but you can for example try the following:
{
Print(IntegerToString(i));
}
When i run this code the log starts at cycle 749
Now if i add a slight delay like this
{
Print(IntegerToString(i));
Sleep(10);
}
Then it does go all the way.
I am not saying that this is the problem but it could be and so it's easy for you to try it by adding a slight delay to your code.
Ok so know that if you call the Print() function too fast repeatedly, especially in fast loops, it can start to skip lines in the log.
I am not sure if that is the problem here but i see a while loop with a print() call in your code.
I do not know how many cycles you loop has but you can for example try the following:
{
Print(IntegerToString(i));
}
When i run this code the log starts at cycle 749
Now if i add a slight delay like this
{
Print(IntegerToString(i));
Sleep(10);
}
Then it does go all the way.
I am not saying that this is the problem but it could be and so it's easy for you to try it by adding a slight delay to your code.
Yes you are right as always.