I initially (asked for help) and wrote a BASIC program in the 6502 pet emulator which added two n-byte integers. However, my feedback was that it was simply adding two 16 bit integers (not adding n-byte integers).
Can anyone help me understand this feedback by looking at my code and point me in the right direction to make a program that adds two n-byte integers?
Thank You for the collaboration!
Documentation: Adds two n-byte integers using absolute indexed addressing. The addends begin at memory locations $0600, $0700 and the answer is at $0800. Byte length of the integers is at $0600 (¢ —> 256)
Machine Code:
18 a2 00 ac 00 06 bd 00
07 7d 00 08 9d 00 09 e8
00 88 00 d0
Op Codes, Documentation, Variables:
A1 = $0600
B1 = $0700
B2 = $0800
Z1 = $0900
[START] = $0500
CLC 18 // loads x with 0
LDX A2 00 // loads length on Y
LDY A1 AC 00 06 // load first operand
loop: LDA B1, x BD 00 07 // adds second operand
ADC B2, x 7D 00 08 // store result
STA Z1, x 9D 00 09 // go to next byte
INX E8 00 // count how many are left
DEY 88 00 // do more if needed
BNE loop D0
It looked to me like your code does what you claim -- adds two N byte operands in little-endian byte order. I vaguely remembered the various addressing modes of the 6502 from my misspent youth and the code seems fine. X is used to index the current byte from the two numbers, Y is a counter for the length of the operands in bytes and you loop over those bytes, stored at addresses 0x0700 and 0x0800 and write the result at address 0x0900.
Rather than get the Commodore 64 out of the attic and try it out I used an online virtual 6502 simulator. On this site we can set the memory address and load the byte values in. They even link to a page to assemble opcodes too. So setting the memory locations 0x0600 to "04" and both 0x0700 and 0x0800 to "04 03 02 01" we should see this code add these two 32 bit values (0x01020304 + 0x01020304 == 0x02040608).
Stepping through the code by clicking on the PC register and setting it to 0x0500 and then single stepping we see there is a bug in your machine code. After INX which compiles to E8 we hit a spurious 0x00 value(BRK) which terminates. The corrected code as below runs to completion and the expected value is seen by reading the memory at 0x0900.
0000 CLC 18
0001 LDX #$00 A2 00
0003 LDY $0600 AC 00 06
0006 LOOP: LDA $0700,X BD 00 07
0009 ADC $0800,X 7D 00 08
000C STA $0900,X 9D 00 09
000F INX E8
0010 DEY 88
0011 BNE LOOP: D0 F3
Memory dump:
:0900 08 06 04 02 00 00 00 00
:0908 00 00 00 00 00 00 00 00