might be the cause of my misunderstanding .. basically when I do sysdev I have to worry about how numbers are stored in RAM if I access memory locations directly (like in my example) .. that is the "core" problem of the whole discussion, as I consider "register" a multibyte memory location...

 

this is 259 (dec) = 0010 0101 1001 (bcd)

so in order to put it in 24bit register i need to :
000000000000001001011001

am i correct??

 

Your first question was to show it as binary .. BCD is binary coded decimal that is not same so double check if you need number with base 2, or bcd coded number, with BCD you need to know do you want packed or not BCD code, your answer is true if you want packed BCD (the most common BCD variant).

note that if you do packed BCD addition, you add the numbers as they were binary numbers (perform normal binary addition) and then add 6 to each digit that is larger then 9

 

Endianness - Wikipedia, the free encyclopedia

that's the answer to my problem (where is that banging head smiley) ..
for some strange reason, I decided to believe that multibyte value is stored in register the same way as it is stored in memory .. hence all the ping pong ... but now i understand it is not

thanks for clarification and sorry for the stubbornest, I had to get to the bottom of it to be able to sleep