I've been using my assembler [1] to write silly little programs for the Color Computer [2] (simulated [3]—easier than setting up my Color Computer, the first computer I owned as a kid). It took me entirely too long to locate a bug in a maze-drawing program I've been writing, and I want to do a deep dive on this.
The program is simple, it just draws a maze, following a few simple rules:
Nothing hard about it, yet the program kept getting stuck.
[Image of a partially drawn maze] [4]
It starts in the upper left, meanders a bit (in blue), backtracks a bit (in red) until it gets stuck. I would then stare at the code until blood formed on my brow, simplify the code if I could, and try again only to watch it get stuck, again.
The issue is that I have no debugger on this system. I don't have two screens upon which to show debugging output. I have no way of single-stepping though the code. I don't even have a log to go through. Debugging consists of running the code, then thinking real hard. And yes, it was a stupid mistake that took all of two lines of code to fix.
[Image of a fully drawn maze] [5]
Now, the question I want to ask is—would I have saved time if I did “unit testing?” Not just testing, which I was doing all along, but the typical style of “unit testing” where you test a “unit” of code (getting back to that question again [6]). Looking over the code [7] (and that version has the bug—see if you can find it; you'll also need these two [8] files [9]), the most isolated “unit” is random (the last subroutine in the code).
;*********************************************************************** ; RANDOM Generate a random number ;Entry: none ;Exit: B - random number (1 - 255) ;*********************************************************************** random ldb lfsr andb #1 negb andb #$B4 stb ,-s ; lsb = -(lfsr & 1) & taps ldb lfsr lsrb ; lfsr >>= 1 eorb ,s+ ; lfsr ^= lsb stb lfsr rts
It implements a linear-feedback shift register [10] with a period of 255 (in that it will return each value in the range of 1‥255 once in some randomesque order before repeating, which is Good Enough™ for this code). So what if we want to test that the code is correct?
And it's here I came to a realization—“unit testing” really depends upon the language and tooling around it. Modern orthodoxy holds “unit testing über alles” and therefore, modern languages and tooling is created to support “unit testing über alles” (and woe betide those who question it). I think my struggle with “unit testing” is that the environments I find myself in don't lend themselves very well to “unit testing.” Even when I worked at The Enterprise, we were using C (C99 at best) and C++ (C++98, maybe C++03?) which take a lot of work upfront to support “unit testing” well, and there wasn't a lot of work upfront to support “unit testing” well, and there was a decidely lack of a “unit testing” framework. And here, there's definitely not a “unit testing” framework. Any “unit testing” I have to do involves writing yet more code. So let's write yet more code and test this sucker.
CHROUT equ $A002 ; BASIC character output routine lfsr equ $F6 ; unused location in direct page org $4000 start ldx #result_array ; point to memory clra ; storing 0 to memory clrb ; 256 bytes to clear .clrarray sta ,x+ ; clear memory decb ; decrement count bne .clrarray ; keep going until count = 0 ldx #result_array ; point to array lda #255 ; cycle length checkrnd bsr random ; get random number in B tst b,x ; have we seen this number? bne .failed ; if so, we have failed inc b,x ; set flag for this number deca ; are we done with the cycle bne checkrnd ; if not, keep going ldx #msg.success ; SUCCESS! bsr puts rts ;--------------------------------------------------- ; Store count (register A) and random # (register B) ; so we can use PEEK in BASIC to see where we failed ;--------------------------------------------------- .failed std lfsr+1 ldx #msg.failed ; failed message bsr puts ; display it rts ; return to BASIC puts.10 jsr [CHROUT] ; print character puts lda ,x+ ; get character bne puts.10 ; if not NUL, print it rts ; return to BASIC ;****************************************************************** random ldb lfsr andb #1 negb andb #$B4 stb ,-s ; lsb = -(lfsr & 1) & taps ldb lfsr lsrb ; lfsr >>= 1 eorb ,s+ ; lfsr ^= lsb stb lfsr rts ;******************************************************************* msg.success asciiz 'SUCCESS!\r' msg.failed asciiz 'FAILED!\r' result_array equ * end start
The tooling here doesn't support linking 6809 code, and I'd rather not have to keep each routine in its own file since the program is so simple and makes editing it easier if everything is in one file (no IDE (Integrated Development Environment) here—and yes, I have thoughts about testing and IDEs but that's beyond the scope for this post). So I have to copy the routine to the test program.
This was something I kept trying to tell my manager at The Enterprise—the test program itself might be buggy (he personally treated the output as gospel—sigh). And the “unit testing” proponents seem to hem and haw about testing the testing code, implying that one does not simply test the testing code. But if programmers aren't trusted to write code and must test, then why are they trusted to write testing code without testing?
It might seem I digress, but I'm not. There are four bugs in the above test. The code I'm testing, random? It was fine. And I wasn't intending to write a buggy test harness, it just happened as I was writing it. Bug one—I forgot to test that random didn't return 0 (that's a degenerate case with LFSR (Linear-Feedback Shift Register)s). Second bug—I forgot to ininitialize the LFSR state with a non-zero value, so random would return nothing but zero. The third bug was thinking I had used the wrong condition when branching to the failure case, but no, I had it right the first time (the fact that I changed it, and then changed it back, is the bug). The actual bug that caused this was the fourth bug, but I have to digress a bit to explain it.
The 6809 has an extensive indexing addressing mode for an 8-bit CPU (Central Processing Unit). One of the modes allow one to use an accumulator register (A, B or D) as an offset to the index register. I used the B register, which contains the random number, as an offset into a 256-element array to track the return values, thus the use of b,x in the code above. What I forgot about in the moment of writing the code is that the “accumulator,index-register” indexing mode sign extends the accumulator. And the first value from random is, due to the LFSR I'm using, if treated as signed, a negative value—it would fail on the very first attempt.
Sigh.
This is why I panicked and thought I botched the conditional branch.
Now, all of that just to test the most isolated of subroutines in the program.
But had I continued, would any form of “unit testing” been beneficial? There's the subroutine point_addr—which converts an X,Y position into a byte address in the frame buffer, and the pixel in said byte. I could have done an exhaustive test of all 4,096 points, again, that's code I would have write (in 6809 Assembly code) and unfortunately, test, to have any confidence in it. And working up the chain, there's getpixel and setpixel. Testing those would require a bit of thought—let's see … getpixel returns the color of the pixel at the given X,Y location on the screen. and assuming point_addr is working, it would only take four tests (one per pixel in the byte) but at this point, would I even trust myself to write the test code?
In fact, would “unit testing” have saved me any time?
Given that I would have to write the testing framework, no, I don't think I would have saved time. Perhaps if I thought the issue through before diving into changing the code, I would have solved this earlier.
And the clues were there. I did discover pretty early on that the bug was in the backtracking code. The top level code is pretty easy to visually inspect:
backtrack lda #BACKTRACK sta color .loop ldd xpos ; check to see if we're back cmpd xstart ; at the starting point, beq done ; and if so, we're done ldd xpos ; can we backtrack NORTH? decb lbsr getpixel cmpb #EXPLORE bne .check_east lbsr move_north.now ; if so, move NORTH and see if bra .probe ; we have to keep backtracking .check_east ldd xpos ; east ... inca lbsr getpixel cmpb #EXPLORE bne .check_west lbsr move_east.now bra .probe .check_west ldd xpos ; yada yada ... deca lbsr getpixel cmpb #EXPLORE bne .check_south lbsr move_west.now bra .probe .check_south ldd xpos incb lbsr getpixel cmpb #EXPLORE bne .probe lbsr move_south.now .probe bsr boxed_in ; can we stop backtracking? bne explore ; if so, go back to exploring bra .loop ; else backtrack some more
The thing to keep in mind here is that the D register is a 16-bit register where the upper 8-bits is the A register, and the lower 8-bits are the B register, and that the 6809 is big-endian. So when we do ldd xpos we are loading the A register with the X coordinate, and B with the Y coordinate. And the move_*.now subroutines work, or else we wouldn't get the starting maze segments at all. So it's clear that setpixel works fine.
The code is getting stuck trying to backtrack along already drawn segments, and it does that by calling getpixel, therefore, it seems prudent to check getpixel. And sure enough, that is where the bug resides.
;************************************************************************* ; GETPIXEL Get the color of a given pixel ;Entry: A - x pos ; B - y pos ;Exit: X - video address ; A - 0 ; B - color ;************************************************************************* getpixel bsr point_addr ; get video address comb ; reverse mask (since we're reading stb ,-s ; the screen, not writing it) ldb ,x ; get video data andb ,s+ ; mask off the pixel .rotate lsrb ; shift color bits deca bne .rotate .done rts ; return color in B ;************************************************************************* ; POINT_ADDR calculate the address of a pixel ;Entry: A - xpos ; B - ypos ;Exit: X - video address ; A - shift value ; B - mask ;************************************************************************* point_addr.bits fcb %00111111,%11001111,%11110011,%11111100 ; masks fcb 6,4,2,0 ; bit shift counts
I've included a bit of point_addr to give some context. point_addr returns the number of shifts required to move the color value into place, and one of those shift values is 0. But getpixel doesn't check to see it's 0 before decrementing it. And thus, getpixel will return the wrong value for the last pixel in any byte. The fix is simple:
tsta beq .done
just before the .rotate label fixes the bug (two instructions, three bytes and here's a link to the fixed code [11]).
Yes, I freely admit that a “unit test“ of this subroutine would have shown the bug. But how much time would I have spent writing the test code to begin with? The only reason it took me as long as it did to find was because the reference code I was using was quite convoluted, and I spent time simplifying the code as I went along (which is worthy of doing anyway).
What I wish the “unit testing” proponents would realize is that easy testing depends upon the language and tooling involved in the project, and what a “unit” is truly depends upon the language. I suspect that “unit test” proponents also find “unit testing” easier to deal with than “integration testing” or even “end-to-end testing,” thus why we get “unit tests über alles” shouted from the roof tops.
[2] https://en.wikipedia.org/wiki/TRS-80_Color_Computer
[3] http://www.6809.org.uk/xroar/
[4] /boston/2023/11/27/stuck-maze.png
[5] /boston/2023/11/27/maze.png
[7] /boston/2023/11/27/maze-bug.asm
[8] /boston/2023/11/27/Coco-DP.i
[9] /boston/2023/11/27/Coco-video.i
[10] https://en.wikipedia.org/wiki/Linear-feedback_shift_register
[11] /boston/2023/11/27/maze.asm
Unit testing on an 8-bit CPU - ZeroBytes
Unit testing on an 8-bit CPU - derp.foo
Unit testing on an 8-bit CPU | Hacker News
![[4] /boston/2023/11/27/stuck-maze.png](/gemini.conman.org/boston/2023/11/27/stuck-maze.png){kind=link}
![[5] /boston/2023/11/27/maze.png](/gemini.conman.org/boston/2023/11/27/maze.png){kind=link}