2

I'm working on a specialty hexadecimal editor that includes a Z80 two-byte pointer converter.

The mathematics behind the conversion are like so:

  1. Take the offset that you wish to point to.
  2. Take the last four digits of the offset, and cut off the rest.
  3. If the offset is outside the range &H4000 - &H7FFF, it must be converted like this: (offset % &H4000) + &H4000. In other words:
    • If the offset is from &H0000 to &H3FFF, add &H4000 to the offset.
    • If the offset is from &H4000 to &H7FFF, do not do anything to the offset.
    • If the offset is from &H8000 to &HBFFF, subtract &H4000 from the offset.
    • If the offset is from &HC000 to &HFFFF, subtract &H8000 from the offset.

My problem is I don't know how I could turn a 5 or 6-digit hex offset into a two-digit offset. How would I shave off the extra bytes at the beginning (step two)?

AlexTheRose
  • 125
  • 1
  • 1
  • 8

2 Answers2

4

With the "remainder" operator, spelled Mod in Visual Basic:

offset Mod &H10000
Kerrek SB
  • 464,522
  • 92
  • 875
  • 1,084
3

The answer using the 'Mod' operator is correct.

But, to be pedantic, this is a 'modulus' operator, and not a 'remainder' operator. There is a difference for negative numbers. (I appreciate we are not talking about negative numbers here.)

See What's the difference between “mod” and “remainder” ？

Community
  • 1
  • 1
debater
  • 466
  • 6
  • 11
  • 3
    Actually the linked MSDN definition of Mod says, surprisingly, that it is the remainder, but then does not discuss negative numbers. Good point to bring it up -- though your answer is probably better suited to be a comment. – Ben Flynn Apr 06 '13 at 21:55