68000 Assembly Language - CMPI.B -

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  What is the content of CCR and D3 after the sequence of following commands? Calculate by hand and show off your work. MOVE.B # 7, D3 CMPI.B # 11, D3

I know that the content of the D3 register will remain unchanged, I do not know how to calculate the CCR Flags

Can you tell me how you did it and which flags were closed and why I really have a hard time understanding it.

D3 CMPI , But surely its less byte will be MOVE due to 7 . (Thanks to @Windand to indicate this.)

The directive set reference will tell you that CMPI works by subtracting operands before the second. It also says that the X flag is not affected, and the other results are set as follows:

  • N : 7 -11 & lt; 0 , therefore N = 1
  • Z : 7-11! = 0 , therefore Z = 0
  • V : 7-11 = -4 , that number For the signed numbers, V = 0 :
  • C : 7-11 = -4 , Which is outside unsigned numbers, C = 1 .

This is the human version for the flag, the CPU actually uses bitrift logic:

  • N : the result of In the supplement represenation of the most important bit 2, the negative number is set to MSB.
  • Z : All the result bits are just a bitword nand Z = 1 if all the bits are zero.
  • V : This is a difficult one. The result is MSB with V = (- R7) * (- A7) * B7 + R7 * A7 * (- B7) , R7 , A7 < MSB of two operands means that you have a signed overflow if subtracting a negative number from a positive number gives you a negative number, or if negative number reduces negatively So lets give you positive.
  • C : the last move of subtraction, aka 9V bit.

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