Slight confusion on how to set var

DRMorrison

I am having a slight issue with setting a var in my main program from and object. I would like to change the var X from within
an object to affect the X var in the main program. I need to use another cog to change the var X, and then when the main program
uses it during its next iteration, it will see the new value that has been changed by the object. I think I need to use the @ symbol somehow,
but I don't fully understand the syntax. Do I simply write @X in the object?

Thanks, Daniel
(the issue is my lack of understanding, or ignorance. I've tried reading about it, but I just don't grasp it quite well enough.)

Mike Green

You have to pass the address of the variable in your main program to your object. Your object is going to have some sort of initialization method like:

pub start(a,b,c)
   ... stuff ...

You need to have another parameter like:

var long addr    { place to save passed address }
pub start(a,b,c,d)
    addr := d   { save the address passed in }

Wherever you want to change the variable, you'd use the following. If the variable is a byte or word, you'd use BYTE() or WORD() instead of LONG():

LONG(addr) := value

In the main program, you'd pass the address of the variable like this:

start(5, 10, 103, @variable)

DRMorrison

So, object looks like...

PUB start(@X)
 WORD(X) := y + z

And then the main program would see the new value the next time it went to use it?
Do I have to call the object from main, or will the value just change from within the object due to
internal processing within the object?

DRMorrison

I think I may have this...
MAIN PROGRAM:

VAR WORD X

OBJ newX: "Xchanger"

PUB Main

 newX.Start(@X)

OBJECT Xchanger:

VAR WORD Xadd

PUB Start(X)
 Xadd := X

PUB xchange

 WORD(Xadd) := {some new value based upon an input etc.}

Does this look correct?
Daniel

msrobots

yes

Mike

DRMorrison

The line:
WORD(Xadd) won't compile, but this will: WORD[Xadd]
I'm using BST

Daniel

Mike Green

Sorry. I got carried away with the parentheses. It indeed is WORD[Xadd]. Looks good.

DRMorrison

Thanks to everyone, Daniel