Bitwise - how to check if a permission is not set?

Hi there,
I'm struggling a bit right now with bitwise operators. I do understand e.g. how to figure out if a permission is set, but how do I check if a permission is not set??

Example:

integer perm_mask = llGetObjectPermMask( MASK_NEXT);
if(perm_mask & PERM_COPY) llSay(0,"The next owner will have copy permissions.");

Well there is nothing like perm_mask &! PERM_COPY of course, and I cannot do a ~ on it either.

???

thx and greetz

A8

I figured it out myself Love those situations, just if someone wants to know it:

integer perm_mask = llGetObjectPermMask( MASK_NEXT);
if((perm & PERM_COPY) != PERM_COPY) llSay(0,"The next owner will have no copy permissions.");

Sure there is. You gave the simplest answer already. I've tested it and used it before and it works well:

integer perm_mask = llGetObjectPermMask(MASK_NEXT);
if(~perm_mask & PERM_COPY) llSay(0, "Sorry, no copy perms for the next owner!";

What happens is the permissions mask gets bitwise inverted, then you apply the AND mask which will now pull out FALSE permissions. You can also squash it all into one if statement which saves a few bytes...

if (~llGetObjectPermMask(MASK_NEXT) & PERM_COPY) llSay (0, "Sorry, no copy perms for the next owner!";

this can be used to advantage sometimes...

if (~llGetObjectPermMask(MASK_NEXT) & (PERM_COPY | PERM_TRANSFER)) llSay (0, "The next owner has some rights restricted.";

This will be true if the next owner either cannot transfer, or cannot copy (or both). If the next owner can copy AND transfer, the condition will not be true.