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I'm doing some exam practice questions and the first part involves printing out the hexadecimal number (say xFDO1) as a binary number.

My code prints the opposite and I know I can reverse this order by making another loop and instead of starting with the zero bit start with the n bit and bit-shift n-1 times to get the next bit and so on for all the bits but I am hoping there is a better way!

I also know I can hard code all of the different bits and check those against the number but still not an elegant solution.

Here is my code for reference:

    .ORIG x3000

    LD  R1, binary  ;loads number we wanna use
    LD R2, maskbit    ;starts with 0000 0000 0000 0001
LD R4, counter  
  loop    
AND R3,R3,#0    ;resets R3
AND R3,R1,R2    ;checks if has bit there
    BRz else 
    LD  R0, ascii1
    BR done
  else   
LD  R0, ascii0
  done  
OUT
    ADD R2,R2,R2    ;shift bit one over
ADD R4,R4,#-1   ;decrement counter
    BRzp loop       ;loops if counter not negative
  HALT

   counter .fill #15
   maskbit .fill x0001
   ascii0  .fill x30
   ascii1  .fill x31
   binary  .fill XAF12
    .END

I am hoping to find a better approach.

bob mcgee
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1 Answers1

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No need to do all of that shifting to the number.

You can use the properties of signed numbers to do this. When you load the number you want to print immediately check if it was negative. Remember that negative numbers have the sign bit set.

Here is some pseudocode

  • Load the number
  • If the number is negative print a 1 else print a 0
  • Do this 15 times:
    • Double the number. (Left shift it by 1).
    • If the number is negative print a 1 else print a 0
Brandon
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