Strange output from Hex() Function

[Tyranna]

I do not know what is going on.  Assuming there is some bug in hex.  I use Ptr(MyVar) and everything outputs good.

If I use Hex(MyVar) , the first conversion by Hex() works, then I add 1 to MyVar and the Hex(MyVar) outputs garbage.

Something is happening in the backgroud with this function that I cannot see.  I have spent over 3 days trying to nail this

down with no luck whatsoever.  Maybe its GOD's way of saying, USE THE POINTER FUNCTION!

Edited July 24, 2017 by Tyranna

[jchd]

Until you post a short reproducer script, this will remain a mystery. But once you do, if ever, I bet all will clear up quickly.

 

[Tyranna]

This reads process memory , there is no real way to give you a 'short' reproducer script.

I could give you a barebones version of what this is doing, but you would have to map it to some running process

on your end and test it that way.  Ill attach just in case...

 

 

Script.rar

Edited July 24, 2017 by Tyranna

[Tyranna]

A new development...

Replacing:

 $MEMOUT &= "  " & Hex( $CurrentMemorySearchAddress ) & " | "

with:

 $MEMOUT &= "  " & Hex( Int( $CurrentMemorySearchAddress) ) & " | "

makes the output good again.

Who can tell me why?

 

 

[czardas]

$CurrentMemorySearchAddress would appear not to be an integer data type. Converting the data gives you the output you expect. Whatever process created the variable $CurrentMemorySearchAddress holds all the clues as to how this situation occured. If the variable was a float, then you will get a completely different output:

1 = 1.0
Hex(1) <> Hex(1.0)

Edit: Didn't notice the first page. LOL

Edited July 26, 2017 by czardas

[Tyranna]

This variable is totally contained within the output subroutine. no declaration of any kind anywhere else.

My script passes $MemorySearchAddress to the sub

DisplayMemory( $MEMIN , $MemorySearchAddress )

then the output sub is:

DisplayMemory( $MEMIN , $CurrentSerachAddress )

all instances of $CurrentSearchAddress are in here.

EndFunc

[czardas]

12 minutes ago, Tyranna said:

no declaration of any kind anywhere else

It had to have come from somewhere. You should follow through with @RTFC's advice. Check the variable type with VarGetType(). It may be being corrupted somewhere, or originally different to what you expected.

[Tyranna]

The var is created when passed to the output subroutine.

The variables first instance is a parameter of the DisplayMemory()  function.

Ex:

DisplayMemory( $MEMIN , $CurrentSerachAddress )

all instances of $CurrentSearchAddress are in here.

EndFunc

[czardas]

I'm sorry; I don't understand. The code you have posted will throw errors. You have two instances of $CurrentSerachAddress: one local and the other one global [edit: might have had something to do with it]. There isn't enough information here to solve this.

Edited July 26, 2017 by czardas

[Tyranna]

Nope, it is just as described.

I used that VarGetType()  , and it comes in as a string, and then, when the addition is used on it, it is converted to a Double.

 

[Tyranna]

I wish I could just declare my variables as type

 

[czardas]

Variables are indeed created as empty strings if not assigned any original value (perhaps contraversial). For it to be an integer type, it would have to be assigned an integer value ($eg = 0). This is why Hex produces a different output, but it isn't entirely clear to me how this code (or Hex for that matter) is meant to behave.

[Tyranna]

So, I declared the variable global up top and set it to zero, and it produces garbage still.

 

[Tyranna]

its still coming as a string and then beeing converter to a double

[czardas]

Yeah, Hex sees an empty string as garbage ==> 00000000!

Conversion to double? It seems you have solved the mystery and found the solution already.

[Tyranna]

but i want it to come into my sub as a specific type!

 

[czardas]

Can't you simply add $CurrentSearchAddress = Int($CurrentSearchAddress) as the first line of your DisplayMemory() function?

[Tyranna]

i can, lol. seems a little redundant...

 

[czardas]

Then do that!

[Tyranna]

thx