Help with understanding M68HC11 arrays

Discussion in 'Homework Help' started by bluesilver89, Mar 28, 2010.

  1. bluesilver89

    Thread Starter New Member

    Mar 28, 2010
    1
    0
    Hello! I'm new here, but I was hoping to get some help understanding how to use arrays in the 68HC11 programming language. I have already completed a beginners Java course, so I understand the concept of an array. However, I do not understand how to manipulate them in an HC11 program.

    My homework assignment is to write an HC11 program that will initialize and manipulate an array, then write a series of 3 loop. The 1st copies an array of constants from EEPROM to RAM (which I attempted in the example below), the 2nd loop involves subtracting the first array element from the second, and then storing the result in the second (a[1] = a[1] - a[0]), and repeat for all the elements except for the assignments themselves (so no a[2] = a[2] - a[1], because a[1] is already being used as an assignment). It is assumed that the array contains an even number of elements. The 3rd loop sums the entire array into a single total and stores it into a variable. Each loop should be within the same program, but seperate from each other.

    I was able to somewhat figure out the first loop, but I am unsure how to manipulate the array from there. I learn best from seeing examples, so would it be possible for someone to show me a basic HC11 program that does something with an array (it doesn't have to be based on my homework, I just need to see an example on how to use arrays). Also, if anyone notices any problems with my posted program, can you point them out to me please? I would definately appreciate any help! Thank you for your time!

    Code ( (Unknown Language)):
    1.  
    2. RAM       EQU       0
    3. EEPROM  EQU       $F800                        
    4. array      FCB       7, 24, 9, -12, 63, 87     * array of 6 constants
    5. asize      EQU       6                    * space reserved for array
    6. total      RMB       1                    * total variable for 3rd loop
    7.  
    8.             ORG       RAM
    9. dst        RMB       asize               * space for destination array
    10.  
    11.             ORG       EEPROM
    12. src        FCC       array
    13.             FCB       0                     * terminate array with null
    14.  
    15. start       LDAB     #asize              * load asize in accumulator B
    16.             LDX       #src                 * load EEPROM in x register
    17.             LDY       #dst                * load RAM in y register
    18. loop       LDAA      0,x                  * load first element of array in x
    19.             STAA     0,y                  * store that element in destination
    20.             BEQ       loop2               * if null, loop to 2nd loop
    21.             INX                              * increment x
    22.             INY                              * increment y
    23.  
    24.             DECB                            * decrement 1 from B
    25.             BNE       loop                 * if asize != 0, branch to loop
    26.  
    27. * all this below is my incomplete attempt to manipulate the array for the 2nd and 3rd loops but I don't think I'm doing it correctly. I think I might have to use the 0,x, 1,x etc, but I'm not sure.
    28.  
    29. loop2      LDAA     array[1]
    30.             LDAB     array[0]
    31.             SBA
    32.             STAA    array[1]
    33. loop3      LDAA    array[0]
    34.             LDAB     array[1]
    35.             ABA
    36.             STAA    total
    37.  
    38. done
    39.  
    40. JMP      END      JMP
    41.  
    42.  
    43.  
     
  2. RiJoRI

    Well-Known Member

    Aug 15, 2007
    536
    26
    Consider using pointers in loop2 as you did in loop1. Or even one pointer:
    Code ( (Unknown Language)):
    1.  
    2.   ldaa   ,x
    3.   inx
    4.   adda  ,x
    5.   staa   ,x
    6.   inx
    7.  
    Please note that I am familiar with the 6809, not the 6811. There may be features the '09 had that the '11 doesn't, and vice-versa.

    Another method:
    Code ( (Unknown Language)):
    1.  
    2.   ldx  #00
    3. loop:
    4.   ldaa  array,x
    5.   inx
    6.   adda  array,x
    7.   staa   array,x
    8.   inx
    9.   cpx    DST
    10.   bne    loop
    11.  
    Question: Did the teacher mention what to do when you have an overflow/underflow? E.g., $80 + $82 = $02 (with the Carry bit set). If not, don't worry about it. You'll get to it pretty soon.

    --Rich
     
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