CODEWITHSUNDEEP
assembly68000easy68k
mira
mira

Reputation: 11

Implementing while loop in 68k

Translate the following code into assembly language and display the final value of Y. (2 pts.)

    X = 1;
    Y = 2;
    WHILE(X <= 10 ){
    IF((X%2) != 0)
        Y = Y * X;
    ELSE 
        Y++;
    X++;}
    Output Y

PLEASE LET ME KNOW IF I AM DOING EVERYTHING RIGHT BECAUSE I AM NOT SURE IF MY OUTPUT IS RIGHT. RESULT IS 3712

*-----------------------------------------------------------
* Title      :
* Written by :
* Date       :
* Description:
*-----------------------------------------------------------
    ORG    $1000
START:                  ; first instruction of program

* Put program code here

  MOVE.W  #1,D1           ;PUT 0 IN D1 (X)
  MOVE.W  #2,D2           ;PUT 1 IN D2 (Y)

LOOP CLR.W   D3         ;Find the remainder
     MOVE.W  D1,D3
     DIVU    #2,D3
     SWAP    D3


     CMP     #0,D3      ;Compare remainder with 0
     BNE     NOTEQUAL   ;If NOT equal, then go to NOT equal

     ADD.W   #1,D2      ;Y++
     ADD.W   #1,D1      ;X++

     CMP     #11,D1     ;Compare D1 with 11
     BEQ     DONE       ;If D1 equals 11, break loop.      
     BRA     LOOP

NOTEQUAL 
      MULU.W  D1,D2     ;Multiply D1 and D2 and store it in D2
      ADD.W   #1,D1     ;X++
      CMP     #11,D1    ;Compare D1 with 11
      BEQ     DONE      ;If D1 equals 11, break loop. 
      BRA     LOOP


DONE LEA MESSAGE,A1
     MOVE.W #14,D0
     TRAP #15

     MOVE.W  D2,D1

     MOVE.W #3,D0
     TRAP #15    


    SIMHALT             ; halt simulator

* Put variables and constants here

MESSAGE DC.W    'Y is: ',0

    END    START        ; last line of source

Upvotes: 0

Views: 742

Answers (1)

chtz
chtz

Reputation: 18807

This will result in unexpected results, if the upper word of D3 contains data larger 1.

LOOP CLR.W   D3      ; D3 = [**** 0000]
     MOVE.W  D1,D3   ; D3 = [****    X]
     DIVU    #2,D3   ; if D3.h > 1, division overflows and D3 does not change
     SWAP    D3      ; Lower word is (X%2) or random data

If you really want to use DIVU make sure to clear the long register (with clr.l d3 or actually faster moveq #0,d3). Just clearing the lower word is pointless, if you overwrite that afterwards.

But as @PeterCordes said, DIVU is very inefficient to check the lower bit (I'll not point out other inefficiencies in your code). Better use btst #0,d1 (sets the Z flag if the bit is empty). Alternatives are:

  • and.w the number 1 with the value you want to test. Also sets the Z-flag if the lowest bit is empty
  • Shift the value to the right (sets the C flag, if the lowest bit was set)

Of course, you can also partially unroll the loop, and exploit the fact that (X%2) != 0 is true every other iteration. And you could fully unroll the loop and directly move the result into your target (that's what most optimizing compilers will be able to do: https://godbolt.org/z/VGecJ3).

Upvotes: 3

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