How do I write the equivalent of SETB 53h in 68000 assembly code? SETB 53h is 8051 assembly code.

SETB 53h sets one bit in the user's program memory. But how do I do the same in 68000 assembly language?
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Based on this previous question I found regarding what SETB 53h does on the 8051... appears to me that SETB sets individual bits at absolute RAM address 20h to 2Fh.  Specifically, SETB 53h would set the 4th bit of the 11th byte in this range, or the 4th bit of absolute address 2Bh

I have not been able to find how to do the exact same thing with the 68000.  I have found this...
...but I don't have enough experience to properly interpret the syntax.  But it appears that the 68000 has a BSET command.  As a best guess, I'm thinking the command...
BSET $2B,4
...would set the 4th bit at memory address 2Bh

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today

From novice to tech pro — start learning today.