Combining two integers into a bitfield

I have two different integers I want give as start parameters via llRezObject(). Is it possible to combine the two integers (eg, 25 and 13) into a bitfield so I can give it as the start parameter, so the rezzed object can decode the bitfield to get the two integers? Apologies if this is a dumb question

I know, there's a more elegant way to do this, but you could try this:

Pass 2513 to the rezed object

In the on_rez-event, you simply do this:

(Pseudo code!)

on_rez(integer pass)
{
integer first = pass/1000;
integer second = pass - (first * 1000);
}

So first should be 25 and second should be 13

(Didn't test this...

One way of doing this would be:

default
{
state_entry()
{
integer a = 4848;
integer b = 2345;
integer combined = (a * 3276 + b;
integer a_new = combined / 32768;
integer b_new = combined % 32768;
llOwnerSay ("a_new: " + (string) a_new + " b_new: " + (string) b_new);
}
}

It can take a little getting your head around

The method is to split the 32 bit integer up into smaller integers by bit shifting and masking.

Its worth getting your head around hexadecimal numbers as it makes things a lot easier to represent.

A very simple example:

integer test;

integer out1;
integer out2;
integer MASK1 = 0xFF;
integer MASK2 = 0xFF00;
default
{
state_entry()
{

}

touch_start(integer total_number)
{
test = (MASK1 & 25) + (MASK2 & (13 << );
llSay(0, (string)test);

out1 = MASK1 & test;
out2 = (MASK2 & test) >> 8;
llSay(0, (string)out1 + " " + (string)out2);
}
}

Thanks Resolver, that works a treat.

It sure can! My feeble brain groans under the strain - I just barely follow what is going on, and there is no way I could explain it But, it does exactly what I need, so thank you very much!