CODEWITHSUNDEEP
arraysassembly68000
Brownstown Balla
Brownstown Balla

Reputation: 1

Summing 2-dimensional array in EASy68K assembly

100x100 array A of integers, one byte each, is located at A. Write a program segment to compute the sum of the minor diagonal, i.e.

SUM = ΣA[i,99-i], where i=0...99

This is what I have so far:

LEA A, A0
CLR.B D0
CLR.B D1
ADDA.L #99, D0
ADD.B (A0), D1
ADD.B #1, D0
BEQ Done
ADDA.L #99,A0
BRA loop

Upvotes: 0

Views: 1068

Answers (1)

Mickaël Pointier
Mickaël Pointier

Reputation: 286

There's quite many issues in this code, including (but not limited to):

  • You use 'Loop' and 'Done', but the labels are not shown in the code
  • You are adding 100 bytes in D1, also as a byte, so you are definitely going to overflow on the results (the target of the sum should at least be 16 bit, so .w or .l addressing)
  • I'm perhaps wrong but I think the 'minor diagonal' goes from the bottom left to the upper right, while your code goes from the top left to the bottom right of the array

On the performance side:

  • You should use the 'quick' variant of the 68000 instruction set
  • Decrement and branch as mentioned by JasonD is more efficient than add/beq

Considering the code was close enough from the solution, here is a variant (I did not test, hope it works)

    lea A+99*100,a0     ; Points to the first column of the last row
    moveq #0,d0         ; Start with Sum=0
    moveq #100-1,d1     ; 100 iterations
Loop    
    moveq #0,d2         ; Clear register long
    move.b (a0),d2      ; Read the byte
    add.l d2,d0         ; Long add
    lea -99(a0),a0      ; Move one row up and one column right
    dbra d1,Loop        ; Decrement d1 and branch to Loop until d1 gets negative
Done
    ; d0 now contains the sum

Upvotes: 1

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