notANON

I’ve been trying to figure out how to do this nearly since the beginning.  This program represents a huge milestone for me.  It’s the first program that I have created from scratch that does something sort of useful.

It’s amazing how much debugging can go into something so simple as a hex counter.  At first I had to press “I” three times for the counter to start moving upward.  Turns out I had a few commands out of order.  Also, I had a bug where the counter location in memory would not have been properly reset to zero if R2 didn’t have a residual value in it.  Then I had some other misplaced bytes that were confusing me a bit.  I think I have all the kinks worked out now so I’d like to present my hex counter:

0000     6400    OUT    4        clear hex display
0002     90         GHI     R0    zero the accumulator
0003     B2         PHI    R2      Set hi byte to zero
0004     B3         PHI    R3      Set hi byte to zero
0005     A3         PLO    R3     Set lo byte to zero
0006     F820    LDI                Set D to 20
0008     A2        PLO    R2      Set R2 to 20    
0009     52        STR           Clear 0020 byte
000A     E2        SEX    R2      Set X to R2
000B     3F0B   BN4               Wait for “I” to be pushed
000D     13         INC    R3      Count up
000E     83        GLO    R3      Set D to the current count
000F     52         STR    R2      Store counter value in 0020
0010      64        OUT    4         Put contents of 0020 on display
0011      22         DEC  R2        Keep X pointing at 0020
0012     3712     B4                   Wait for “I” to be lifted
0014     300B    BR                  Loop back to wait for “I”

Please excuse the formatting, it’s late.  Anyhow, the program sets itself up by clearing the display, zeroing R3, setting R2 to 20, clearing the byte at memory location 20 and of course setting X to R2.  R2 does double duty in this program.  It points X at byte 0020 in memory and it also tells the STR instruction where to store the value of D.

After the initial setup, we jump into the loop.  First we increment R3(the counter).  Next we stick the counter value in D so we can then store D at memory location 0020.  Then we display the value of 0020 since that is where R2 is pointing.  The 64 OUT 4 instruction incremented the value in R2 so we decrement it immediately so R2 still points at the correct location and doesn’t leak memory.  If you want a good demo of a memory leak, change the value of 0011 from 22 to C4 and then run the program.

I know this program isn’t rocket science but give me a break, I’m still in chapter 4.  Looks like I still have a LONG ways to go to program in “hello world” even…