Question about expression search

[fabsyi]

When doing an expression search, is it possible to access the address of the current value?

This would be useful in doing something like:
??==10 && [&(??)+10]==11

A search that looks at a known offset from the current position.


Also, another question in regard to the Hex Editor. Locations in the hex editor change colour when a game changes their value. Is there a way to 'refresh' the view then, so that the coloured bytes are reset, so that it is easy to see future RAM changes by the game, without seeing those that changed in a past test.

Thanks for any replies.

[L. Spiro]

#1: There is not, but I plan to add an option to treat ?? as either the value or the address.

I originally built it so that if you set it as a Pointer it was the address. But every search I did had this in it: [??]. An immediate dereference of the address. This slowed the search down tons since it would manually dereference every address, even after the scanner already buffered all of the values.

So the alternative is to allow the user to switch between them. For now it only allows the faster and more common approach.

Then your example would be:
[??] == 10 && [??+10] == 11



#2: I never got around to adding that feature back.
I currently do this by loading a new Hex Editor tab to the same location.


L. Spiro

[fabsyi]

Thanks for your reply.

I have one further question regarding the sub search to an expression search.
I assume there is no way to query the result of the previous search, in order to do something like an 'is changed' type sub-search. I'm not expecting there to be, just thought I might ask in case you by chance implemented it.

Anyway, thanks for all your good work on this program, it's so useful in many different applications.

[L. Spiro]

There is not yet, though I have considered changing the ?? keyword to $$, so I could also use $0, $1, etc. like in mIRC script to add multiple custom variables to expressions.
With that I could provide $$ as the current value (or address) and $0 be the previous.


L. Spiro