Understanding 6502 assembly on the Commodore 64 - (7) The Assembler and Machine code Compiling

So what happened when the compiler ran??  It took our assembly code and converted it into machine code and stored it in a prg File that the commodore could read.  The commodore loaded the file, where it was instructed to, starting at C000.

Lets see what the code looks like, waiting to be executed








Upon being loaded the computer waits with a ready prompt.  You will want to execute the program at its memory location 49152 (HEX $C000), but before we do that lets look at the code in memory

(C:$c051) m c000 c110
>C:c000  a9 93 20 d2  ff a9 01 8d  15 d0 a9 aa  8d 00 d0 a9   .. .............
>C:c010  8f 8d 01 d0  a9 0b 8d f8  07 a2 00 20  22 c0 20 2e   ........... ". .
>C:c020  c0 60 bd 43  c0 9d c0 02  e8 e0 40 d0  f5 60 a9 01   .`.C......@..`..
>C:c030  8d 10 d0 a9  3b 8d 00 d0  a9 32 8d 01  d0 a9 f2 8d   ....;....2......
>C:c040  27 d0 60 03  ff 00 07 ff  80 0f ff 80  1f 83 80 3e   '.`............>
>C:c050  00 80 7c 00  7f 78 00 7e  f8 00 7c f0  00 78 f0 00   ..|..x.~..|..x..
>C:c060  70 f0 00 00  f0 00 70 f0  00 78 f8 00  7c 78 00 7e   p.....p..x..|x.~
>C:c070  7c 00 7f 3e  00 80 1f 83  80 0f ff 80  07 ff 80 03   |..>............
>C:c080  ff 00 00 00  00 00 00 00  00 00 00 00  00 00 00 00   ................

 What we see here is a collection of OPCODES and values

The OPCODE is a processor instruction, each one may include zero or several values after it, depending on the situation.




Lets look at the code highlighted , Red are OPCODES green are values


(C:$c051) m c000 c110
>C:c000  a9 93 20 d2  ff a9 01 8d  15 d0 a9 aa  8d 00 d0 a9   .. .............
>C:c010  8f 8d 01 d0  a9 0b 8d f8  07 a2 00 20  22 c0 20 2e   ........... ". .
>C:c020  c0 60 bd 43  c0 9d c0 02  e8 e0 40 d0  f5 60 a9 01   .`.C......@..`..
>C:c030  8d 10 d0 a9  3b 8d 00 d0  a9 32 8d 01  d0 a9 f2 8d   ....;....2......
>C:c040  27 d0 60 03  ff 00 07 ff  80 0f ff 80  1f 83 80 3e   '.`............>
>C:c050  00 80 7c 00  7f 78 00 7e  f8 00 7c f0  00 78 f0 00   ..|..x.~..|..x..
>C:c060  70 f0 00 00  f0 00 70 f0  00 78 f8 00  7c 78 00 7e   p.....p..x..|x.~
>C:c070  7c 00 7f 3e  00 80 1f 83  80 0f ff 80  07 ff 80 03   |..>............
>C:c080  ff 00 00 00  00 00 00 00  00 00 00 00  00 00 00 00   ................



We start with


[A9] - Load accumulator , with the following value 93

[20] - Jump to Sub Routine, followed by D2 and FF.   The 6502 is Little Endian and writes in memory addresses backwards. It is actually FF D2, our CHROUT routine, and comes back to the next instruction after the JSR

[A9] - Load accumulator , with the following value 01

[8D] - Store Accumulator to 15 D0, which we know as $D015

[A9] - Load accumulator , with the following value AA

[8D] - Store Accumulator to 00 D0, which we know as $D000

[A9] - Load accumulator , with the following value 8F

[8D] - Store Accumulator to 01 D0, which we know as $D001

[A9] - Load accumulator , with the following value 0B

[8D] - Store Accumulator to F8 07, which we know as $07F8

[A2] - Load X register , with the following value 00

[20] - Jump to Sub Routine to 22 C0, which we know as $C022 (SPR0LOADLOOP)

[20] - Jump to Sub Routine to 2E C0, which we know as $C02E (MOVEANDCOLOR)

[60] - Return from Sub Routine

[BD] - Load accumulator ABSOLUTE,X with the following value C043 the beginning of sprite data

[9D] - Store accumulator ABSOLUTE,X to the area 02C0 the beginning of sprite 64k block

[E8] - Increment X  (X=X+1)

[E0] - Compare X to hex 40 

[D0] - Branch on not equal, to relative address F5 known as -11(decimal) $C022 from current PC

[60] - Return from Sub Routine

[A9] - Load accumulator , with the following value 01

[8D] - Store Accumulator to 10 D0, which we know as $D010

[A9] - Load accumulator , with the following value 3B

[8D] - Store Accumulator to 00 D0, which we know as $D000

[A9] - Load accumulator , with the following value 32

[8D] - Store Accumulator to 01 D0, which we know as $D001

[A9] - Load accumulator , with the following value F2

[8D] - Store Accumulator to  27 D0, which we know as $D027

[60] - Return from Sub Routine

See the highlighted blue segment, thats out sprite0 data directly following the RTS instruction [60] Starting at $C041to $C082


If we break this down any further, we will literally be left with Binary Machine code of On's and OFFs

Just to display this example, the final code that the machine would read would look like this, I adding the HEX above for view ability, showing the first 5 bytes of the program

     A9              93            20             D2          FF
10101001 10010011 00100000 11010010 11111111


NEXT----->
Understanding 6502 assembly on the Commodore 64 - (8) Inputs in Assembly

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