CPA in assembly language. How CPA and Accumulator works?

Asked Oct 12 '18 at 05:24

Active Oct 12 '18 at 05:27

Viewed 291 times

According to the book , In the assembly language, opcode CPA compares the operand in the memory and the data in the accumulator. I want to know how the instruction CPA affects A register(accumulator)

BR main 

NUM1: .block 2

NUM2: .block 2

MSG1: .ascii “A register is less than 0\0x00”

MSG2: .asciiA register is more than 0\0x00”

Compared: .block2 

main: 

      DECI NUM1,d.    ; for example, 1

      DECI NUM2,d.          ; for example, 4

      LDA NUM1,d

      CPA NUM2,d.      ;after comparing num1(which is now in - 
                    -Accumulator) and num 2(in the memory)


      STA Compared,d.  ;if i store the contents in the Accumulator,
                  Whih figure would accumulator have? 3 or 2?


      BRlT lessThan 
      BR   End

LessThan: 

      STRO msg1,d

      DECO compared,d

          BR END


END: 

      stop 

      .End

According to PEP/8 simulator, The result was “the register is less than 0”, while the accululator is still 1.

What happens when using CPA? Please help me ((

edited Oct 12 '18 at 05:27

Sami Kuhmonen

30,146
9
61
74

asked Oct 12 '18 at 05:24

AssemblePlease

Comparison is comparison, it doesn’t change the register. – Sami Kuhmonen Oct 12 '18 at 05:28
Then the book Computer Science Illuminated 6th edition might be wrong because it says ‘Mnemonic BRLT means ‘Set the Program Counter to thre operand if the Accumulator is less than zero’ – AssemblePlease Oct 12 '18 at 07:04
#Sami Kuhmonen, and thank you for your quick response)) – AssemblePlease Oct 12 '18 at 07:05

CPA in assembly language. How CPA and Accumulator works?

0 Answers0