Clarification on Ctrade PositionOpen

#1

Hello!

I am a bit confused about the volume param in the PositionOpen function https://www.mql5.com/en/docs/standardlibrary/tradeclasses/ctrade/ctradepositionopen

The function definition does not give you any info on it, so I do not know what the volume is based on but I am guessing that it means how many lots, so 1 is one lot. And if that is the case, is there a way of finding out the lot size programmatically? Also, is it possible to open positions with a fraction of a lot like 0.7 ?

Ideally what I am really looking for is a way to open positions based on how much money I want to invest and with how much margin in the simplest way possible.

Thanks in advance for the help.

#2

Yes volume is always “how many lots”.

The value you can use depends of the symbol specifications. Please check the documentation.

#3

Thank you, I found the link very useful. Also is the lot size equal to SYMBOL_TRADE_CONTRACT_SIZE ?

#4

That’s the size of 1 lot.

Example: EURUSD, contract size is 100,000 (1 standard lot).

Volume = 0.7 so that means you will trade 70,000 EUR

#5

Thank you very much! Btw I saw that you live in Belgium and I am currently living in Brussels, let me know if you want to have a beer sometime, I´ll pay of course, given that you are the one answering to all my posts I think is the least I could do :slight_smile:

#6

Maybe one day ;-)…