After a couple of intense days of hacking, this code compiles and runs:
Choice = ' '
while Choice != '':
try:
print()
print('Your choices are:')
print(' 0 - divide by zero')
print()
Choice = input('So which is it? ')
if Choice == '0':
A = 0
print(6//A)
elif Choice != '':
print('I do not understand.')
except:
print('Caught ZeroDivisionError')There is still a long way to go, but that was a major step.
A peek under the hood…
This is a skeleton program:
try:
# main code suite
pass
except:
# except code suite
pass
else:
# else code suite
pass
finally:
# finally code suite
passThis is the raw assembly listing:
00177 ; 00001 try:
023B 20 0694 [6] 00178 jsr PushExceptionContext
023E A9 4C [2] 00179 lda #L00000&$FF
0240 85 4F [3] 00180 sta ExceptionHandler
0242 A9 02 [2] 00181 lda #L00000>>8
0244 85 50 [3] 00182 sta ExceptionHandler+1
00183 ; 00002 # main code suite
00184 ; 00003 pass
00185 ; 00004 except:
0246 20 06AE [6] 00186 jsr PopExceptionContext
0249 4C 0255 [3] 00187 jmp L00001
024C 00188 L00000
024C 20 06AE [6] 00189 jsr PopExceptionContext
00190 ; 00005 # except code suite
00191 ; 00006 pass
00192 ; 00007 else:
024F 20 025B [6] 00193 jsr L00002
0252 4C 025C [3] 00194 jmp L00003
0255 00195 L00001
00196 ; 00008 # else code suite
00197 ; 00009 pass
00198 ; 00010 finally:
0255 20 025B [6] 00199 jsr L00002
0258 4C 025C [3] 00200 jmp L00003
025B 00201 L00002
00202 ; 00011 # finally code suite
00203 ; 00012 pass
025B 60 [6] 00204 rts
025C 00205 L00003This is the listing rearranged slightly to put the lines of Python source as comments near the generated snippets:
00177 ; 00001 try:
023B 20 0694 [6] 00178 jsr PushExceptionContext
023E A9 4C [2] 00179 lda #L00000&$FF
0240 85 4F [3] 00180 sta ExceptionHandler
0242 A9 02 [2] 00181 lda #L00000>>8
0244 85 50 [3] 00182 sta ExceptionHandler+1
00183 ; 00002 # main code suite
00184 ; 00003 pass
0246 20 06AE [6] 00186 jsr PopExceptionContext
0249 4C 0255 [3] 00187 jmp L00001
00185 ; 00004 except:
024C 00188 L00000
024C 20 06AE [6] 00189 jsr PopExceptionContext
00190 ; 00005 # except code suite
00191 ; 00006 pass
024F 20 025B [6] 00193 jsr L00002
0252 4C 025C [3] 00194 jmp L00003
00192 ; 00007 else:
0255 00195 L00001
00196 ; 00008 # else code suite
00197 ; 00009 pass
0255 20 025B [6] 00199 jsr L00002
0258 4C 025C [3] 00200 jmp L00003
00198 ; 00010 finally:
025B 00201 L00002
00202 ; 00011 # finally code suite
00203 ; 00012 pass
025B 60 [6] 00204 rts
025C 00205 L00003The compiler is a single-pass recursive descent parser doing syntax-directed code generation. Things like an empty else or finally part ought to be optimized out, but that is difficult to do without multiple passes, building a parse tree and using that to generate code or implementing a post-compile code optimizer.
Last time, the code generated for try/except/else/finally was presented. Here it is again, but with some minor edits of the Python source code to reduce the number of lines and also with Python code comments moved in each assembly language source file for better correlation to the relevant machine code.
try:
pass # main code suite
except:
pass # except code suite
else:
pass # else code suite
finally:
pass # finally code suite 00177 ; 00001 try:
023B 20 0691 [6] 00178 jsr PushExceptionContext
023E A9 4C [2] 00179 lda #L00000&$FF
0240 85 4F [3] 00180 sta ExceptionHandler
0242 A9 02 [2] 00181 lda #L00000>>8
0244 85 50 [3] 00182 sta ExceptionHandler+1
00183 ; 00002 pass # main code suite
0246 20 06AB [6] 00184 jsr PopExceptionContext
0249 4C 0255 [3] 00185 jmp L00001
00186 ; 00003 except:
024C 00187 L00000
024C 20 06AB [6] 00188 jsr PopExceptionContext
00189 ; 00004 pass # except code suite
024F 20 025B [6] 00190 jsr L00002
0252 4C 025C [3] 00191 jmp L00003
00192 ; 00005 else:
0255 00193 L00001
00194 ; 00006 pass # else code suite
0255 20 025B [6] 00195 jsr L00002
0258 4C 025C [3] 00196 jmp L00003
00197 ; 00007 finally:
025B 00198 L00002
00199 ; 00008 pass # finally code suite
025B 60 [6] 00200 rts
025C 00201 L00003Moving the call of the finally code to the bottom allows saving an instance of the instruction for each except clause
00177 ; 00001 try:
023B 20 068E [6] 00178 jsr PushExceptionContext
023E A9 4C [2] 00179 lda #L00000&$FF
0240 85 4F [3] 00180 sta ExceptionHandler
0242 A9 02 [2] 00181 lda #L00000>>8
0244 85 50 [3] 00182 sta ExceptionHandler+1
00183 ; 00002 pass # main code suite
0246 20 06A8 [6] 00184 jsr PopExceptionContext
0249 4C 0252 [3] 00185 jmp L00001
00186 ; 00003 except:
024C 00187 L00000
024C 20 06A8 [6] 00188 jsr PopExceptionContext
00189 ; 00004 pass # except code suite
024F 4C 0256 [3] 00190 jmp L00003
00191 ; 00005 else:
0252 00192 L00001
00193 ; 00006 pass # else code suite
0252 4C 0256 [3] 00194 jmp L00003
00195 ; 00007 finally:
0255 00196 L00002
00197 ; 00008 pass # finally code suite
0255 60 [6] 00198 rts
0256 00199 L00003
0256 20 0255 [6] 00200 jsr L00002More importantly, this seemingly trivial transformation creates significant opportunities for a dumb single-pass compiler with but one lexical token of lookahead to eliminate unneeded elements.
To begin with, consider the case in which the else clause is not present,
00177 ; 00001 try:
023B 20 06A9 [6] 00178 jsr PushExceptionContext
023E A9 4C [2] 00179 lda #L00000&$FF
0240 85 4F [3] 00180 sta ExceptionHandler
0242 A9 02 [2] 00181 lda #L00000>>8
0244 85 50 [3] 00182 sta ExceptionHandler+1
00183 ; 00002 pass # main code suite
0246 20 06C3 [6] 00184 jsr PopExceptionContext
0249 4C 0253 [3] 00185 jmp L00001
00186 ; 00003 except:
024C 00187 L00000
024C 20 06C3 [6] 00188 jsr PopExceptionContext
00189 ; 00004 pass # except code suite
024F 4C 0253 [3] 00190 jmp L00001
00191 ; 00005 finally:
0252 00192 L00002
00193 ; 00006 pass # finally code suite
0252 60 [6] 00194 rts
0253 00195 L00001
0253 20 0252 [6] 00196 jsr L00002And the case in which the finally clause is not present.
00177 ; 00001 try:
023B 20 06A5 [6] 00178 jsr PushExceptionContext
023E A9 4C [2] 00179 lda #L00000&$FF
0240 85 4F [3] 00180 sta ExceptionHandler
0242 A9 02 [2] 00181 lda #L00000>>8
0244 85 50 [3] 00182 sta ExceptionHandler+1
00183 ; 00002 pass # main code suite
0246 20 06BF [6] 00184 jsr PopExceptionContext
0249 4C 0252 [3] 00185 jmp L00001
00186 ; 00003 except:
024C 00187 L00000
024C 20 06BF [6] 00188 jsr PopExceptionContext
00189 ; 00004 pass # except code suite
024F 4C 0252 [3] 00190 jmp L00003
00191 ; 00005 else:
0252 00192 L00001
00193 ; 00006 pass # else code suite
0252 00194 L00003Note that an empty finally subroutine will be generated if the compiler detects in the try clause a return or a raise statement or a break or continue statement which transfers the flow of control out of the try clause.
Now consider the case in which else and finally are absent. This is one of the two minimal forms of structured exception handling allowed in Python and is perhaps the most common use.
00177 ; 00001 try:
023B 20 06A2 [6] 00178 jsr PushExceptionContext
023E A9 4C [2] 00179 lda #L00000&$FF
0240 85 4F [3] 00180 sta ExceptionHandler
0242 A9 02 [2] 00181 lda #L00000>>8
0244 85 50 [3] 00182 sta ExceptionHandler+1
00183 ; 00002 pass # main code suite
0246 20 06BC [6] 00184 jsr PopExceptionContext
0249 4C 024F [3] 00185 jmp L00001
00186 ; 00003 except:
024C 00187 L00000
024C 20 06BC [6] 00188 jsr PopExceptionContext
00189 ; 00004 pass # except code suite
024F 00190 L00001The other minimal form has only the try and finally clauses.
00177 ; 00001 try:
023B 20 06B2 [6] 00178 jsr PushExceptionContext
023E A9 4C [2] 00179 lda #L00000&$FF
0240 85 4F [3] 00180 sta ExceptionHandler
0242 A9 02 [2] 00181 lda #L00000>>8
0244 85 50 [3] 00182 sta ExceptionHandler+1
00183 ; 00002 pass # main code suite
0246 20 06CC [6] 00184 jsr PopExceptionContext
0249 4C 025A [3] 00185 jmp L00001
024C 00186 L00000
024C 20 06CC [6] 00187 jsr PopExceptionContext
024F 20 0259 [6] 00188 jsr L00002
0252 A6 52 [3] 00189 ldx Exception
0254 A5 53 [3] 00190 lda Exception+1
0256 4C 06AB [3] 00191 jmp Raise
00192 ; 00003 finally:
0259 00193 L00002
00194 ; 00004 pass # finally code suite
0259 60 [6] 00195 rts
025A 00196 L00001
025A 20 0259 [6] 00197 jsr L00002In this case, any exceptions are reraised after the finally clause has been executed.
Not too shabby for a dumb single-pass compiler…
The test code has been slightly restructured to test the compilation of break and continue out of the try clause.
Choice = ' '
while Choice != '':
try:
print()
print('Your choices are:')
print(' 0 - divide by zero')
print()
Choice = input('So which is it? ')
if Choice == '0':
A = 0
print(6//A)
elif Choice == '':
break
print('I do not understand.')
continue
except: # ZeroDivisionError:
print('Caught ZeroDivisionError')
else:
print('In else')
finally:
print('Finally!')A break statement now compiles to this if it is within the try clause.
00374 ; 00014 break
03C5 20 08B2 [6] 00375 jsr PopExceptionContext
03C8 20 043D [6] 00376 jsr L00006
03CB 4C 0461 [3] 00377 jmp L00003However, the break and continue statements in a loop fully contained within a try clause like this one do not invoke the finally clause. That was somewhat tricky to implement.
try:
Choice = ' '
while Choice != '':
print()
print('Your choices are:')
print(' 0 - divide by zero')
print()
Choice = input('So which is it? ')
if Choice == '0':
A = 0
print(6//A)
elif Choice == '':
break
print('I do not understand.')
continue
except: # ZeroDivisionError:
print('Caught ZeroDivisionError')
else:
print('In else')
finally:
print('Finally!')All in all it’s just another brick in the wall.
From the When it Rains, it Pours Department,
try:
pass # main code suite
except:
pass # except code suite
else:
pass # else code suite
finally:
pass # finally code suiteThe documentation does not say that break and continue are not allowed in the except, else and finally clauses.
Compiling these statements within an except or else clause need not (and should not) pop the exception stack. But they should still invoke the finally clause.
Compiling these statements within the finally clause should do neither.
Edit: One final detail. Because it was compiled as a subroutine, any flow of control out of the finally code suite other than returning at the end must remove the return address from the stack. Discovered that one the hard way…
After the extensive work on exception handling, I indulged in a couple of rabbit holes.
The first is writing a “small” emulator to enable running programs written for the SWTPC 6800 on a system based on the 6809. This is possible because a “full boat” 6809 system contains 56K of RAM while the maximum is 40K for the 6800.
That gives me a whopping 16K of memory for the emulator. I map the I/O devices from $80xx on the 6800 to $E0xx on the 6809, and thunk calls to the FLEX operating system from $ADxx to $CDxx.
At this point, I can run the assembler and the BASIC and Extended BASIC interpreters on simple programs. Performance is about 1/10 of native speeds as expected.
The second is a bit more ambitious and has direct application to much of what I have been doing in this thread, writing compilers.
In the early days, I was obsessed with a programming language called PL/M. It was modeled after PL/I, a programming language from the IBM mainframe, but was intended to be used to program microprocessors. It was somewhat legendary because Gary Kildall reportedly used it to write his CP/M operating system.
I never got to program in PL/M. As far as I know, it was never available to actually run on CP/M, but it had to be used either on an expensive development system from Intel or some mini and mainframe computers.
PL/M is a fairly simple language. It has only two data types: byte and address. Address was the term for two-byte quantities. It was actually more of a high-level assembler than C, its minicomputer competitor. Arithmetic is always treated as unsigned. Adding or subtracting byte values result in a byte value. Mixing a byte and an address promotes the byte by zero extension; storing an address into a byte lops off the upper byte.
A great summary is here if you want to see more: https://www.autometer.de/unix4fun/z80pack/doc_cpm_plm.html
So with the technology I already have, I put together a compiler for a subset of PL/M. I targeted the 6800 processor because it is something I know well.
P$Test: do;
declare (A0, A1, A2) address, (B0, B1, B2) byte;
B0 = 1;
B1 = 2 + B0;
B1 = B1 + 1;
B2 = B0 + B1;
A0 = B0;
B2 = 3 + A0;
A0 = B0 + 1;
A0 = B0 + 256;
A0 = B0 + 257;
B2 = A0;
A1 = A0 + B1;
A2 = B2 + A1;
A0 = 1;
A1 = 2 + A1;
A2 = A0 + A1;
A2 = A0 + 0;
A2 = A0 + 1;
A2 = A0 + 256;
A2 = A0 + 257;
A2 = 256 + B0;
A2 = 257 + B0;
A2 = 1 + 2;
goto 0AD03h;
end P$Test;compiles to
00001 * P$Test: do;
00002
00003 lib ptest.dat
. 00004
.0000 00005 _PTEST.A0 rmb 2
. 00006
.0002 00007 _PTEST.A1 rmb 2
. 00008
.0004 00009 _PTEST.A2 rmb 2
. 00010
.0006 00011 _PTEST.B0 rmb 1
. 00012
.0007 00013 _PTEST.B1 rmb 1
. 00014
.0008 00015 _PTEST.B2 rmb 1
00016
0009 00017 _PTEST
00018 *
00019 * declare (A0, A1, A2) address, (B0, B1, B2) byte;
00020 *
00021 * B0 = 1;
0009 C6 01 [2] 00022 ldab #1
000B D7 06 [4] 00023 stab _PTEST.B0
00024 * B1 = 2 + B0;
000D D6 06 [3] 00025 ldab _PTEST.B0
000F CB 02 [2] 00026 addb #2
0011 D7 07 [4] 00027 stab _PTEST.B1
00028 * B1 = B1 + 1;
0013 D6 07 [3] 00029 ldab _PTEST.B1
0015 5C [2] 00030 incb
0016 D7 07 [4] 00031 stab _PTEST.B1
00032 *
00033 * B2 = B0 + B1;
0018 D6 06 [3] 00034 ldab _PTEST.B0
001A DB 07 [3] 00035 addb _PTEST.B1
001C D7 08 [4] 00036 stab _PTEST.B2
00037 *
00038 * A0 = B0;
001E D6 06 [3] 00039 ldab _PTEST.B0
0020 4F [2] 00040 clra
0021 97 00 [4] 00041 staa _PTEST.A0
0023 D7 01 [4] 00042 stab _PTEST.A0+1
00043 *
00044 * B2 = 3 + A0;
0025 D6 01 [3] 00045 ldab _PTEST.A0+1
0027 96 00 [3] 00046 ldaa _PTEST.A0
0029 CB 03 [2] 00047 addb #3
002B 89 00 [2] 00048 adca #0
002D D7 08 [4] 00049 stab _PTEST.B2
00050 *
00051 * A0 = B0 + 1;
002F D6 06 [3] 00052 ldab _PTEST.B0
0031 5C [2] 00053 incb
0032 4F [2] 00054 clra
0033 97 00 [4] 00055 staa _PTEST.A0
0035 D7 01 [4] 00056 stab _PTEST.A0+1
00057 *
00058 * A0 = B0 + 256;
0037 D6 06 [3] 00059 ldab _PTEST.B0
0039 86 01 [2] 00060 ldaa #1
003B 97 00 [4] 00061 staa _PTEST.A0
003D D7 01 [4] 00062 stab _PTEST.A0+1
00063 *
00064 * A0 = B0 + 257;
003F D6 06 [3] 00065 ldab _PTEST.B0
0041 4F [2] 00066 clra
0042 CB 01 [2] 00067 addb #1
0044 89 01 [2] 00068 adca #1
0046 97 00 [4] 00069 staa _PTEST.A0
0048 D7 01 [4] 00070 stab _PTEST.A0+1
00071 *
00072 * B2 = A0;
004A D6 01 [3] 00073 ldab _PTEST.A0+1
004C D7 08 [4] 00074 stab _PTEST.B2
00075 *
00076 * A1 = A0 + B1;
004E D6 01 [3] 00077 ldab _PTEST.A0+1
0050 96 00 [3] 00078 ldaa _PTEST.A0
0052 DB 07 [3] 00079 addb _PTEST.B1
0054 89 00 [2] 00080 adca #0
0056 97 02 [4] 00081 staa _PTEST.A1
0058 D7 03 [4] 00082 stab _PTEST.A1+1
00083 *
00084 * A2 = B2 + A1;
005A D6 08 [3] 00085 ldab _PTEST.B2
005C 4F [2] 00086 clra
005D DB 03 [3] 00087 addb _PTEST.A1+1
005F 99 02 [3] 00088 adca _PTEST.A1
0061 97 04 [4] 00089 staa _PTEST.A2
0063 D7 05 [4] 00090 stab _PTEST.A2+1
00091 *
00092 * A0 = 1;
0065 CE 0001 [3] 00093 ldx #1
0068 DF 00 [5] 00094 stx _PTEST.A0
00095 * A1 = 2 + A1;
006A D6 03 [3] 00096 ldab _PTEST.A1+1
006C 96 02 [3] 00097 ldaa _PTEST.A1
006E CB 02 [2] 00098 addb #2
0070 89 00 [2] 00099 adca #0
0072 97 02 [4] 00100 staa _PTEST.A1
0074 D7 03 [4] 00101 stab _PTEST.A1+1
00102 *
00103 * A2 = A0 + A1;
0076 D6 01 [3] 00104 ldab _PTEST.A0+1
0078 96 00 [3] 00105 ldaa _PTEST.A0
007A DB 03 [3] 00106 addb _PTEST.A1+1
007C 99 02 [3] 00107 adca _PTEST.A1
007E 97 04 [4] 00108 staa _PTEST.A2
0080 D7 05 [4] 00109 stab _PTEST.A2+1
00110 *
00111 * A2 = A0 + 0;
0082 DE 00 [4] 00112 ldx _PTEST.A0
0084 DF 04 [5] 00113 stx _PTEST.A2
00114 *
00115 * A2 = A0 + 1;
0086 D6 01 [3] 00116 ldab _PTEST.A0+1
0088 96 00 [3] 00117 ldaa _PTEST.A0
008A CB 01 [2] 00118 addb #1
008C 89 00 [2] 00119 adca #0
008E 97 04 [4] 00120 staa _PTEST.A2
0090 D7 05 [4] 00121 stab _PTEST.A2+1
00122 *
00123 * A2 = A0 + 256;
0092 D6 01 [3] 00124 ldab _PTEST.A0+1
0094 96 00 [3] 00125 ldaa _PTEST.A0
0096 4C [2] 00126 inca
0097 97 04 [4] 00127 staa _PTEST.A2
0099 D7 05 [4] 00128 stab _PTEST.A2+1
00129 *
00130 * A2 = A0 + 257;
009B D6 01 [3] 00131 ldab _PTEST.A0+1
009D 96 00 [3] 00132 ldaa _PTEST.A0
009F CB 01 [2] 00133 addb #1
00A1 89 01 [2] 00134 adca #1
00A3 97 04 [4] 00135 staa _PTEST.A2
00A5 D7 05 [4] 00136 stab _PTEST.A2+1
00137 *
00138 * A2 = 256 + B0;
00A7 D6 06 [3] 00139 ldab _PTEST.B0
00A9 86 01 [2] 00140 ldaa #1
00AB 97 04 [4] 00141 staa _PTEST.A2
00AD D7 05 [4] 00142 stab _PTEST.A2+1
00143 *
00144 * A2 = 257 + B0;
00AF D6 06 [3] 00145 ldab _PTEST.B0
00B1 4F [2] 00146 clra
00B2 CB 01 [2] 00147 addb #1
00B4 89 01 [2] 00148 adca #1
00B6 97 04 [4] 00149 staa _PTEST.A2
00B8 D7 05 [4] 00150 stab _PTEST.A2+1
00151 *
00152 * A2 = 1 + 2;
00BA CE 0003 [3] 00153 ldx #3
00BD DF 04 [5] 00154 stx _PTEST.A2
00155 *
00156 * goto 0AD03h;
00157
00BF 7E AD03 [3] 00158 jmp $AD03
00159 *
00160 * end P$Test;But wait, there’s more.
PL/M was known for being a fairly good optimizing compiler for its time. My simplistic approach of generating code as I parse will not do.
Something I had always intended to do was to build a tree while parsing to represent the program and use that to generate object code. Separating parsing and code generation results in two moderately complicated pieces instead of a single monstrously complex beast. More importantly, it opens the door to insert an optimizer in between.
The PL/M compiler is simple enough to be a good test platform for building this.
What I have right now is a single-line compiler to convert an assignment statement into a tree, then generate assembly language from that.
00012 **A0 = (B1 + 1) + A1;
00013
00014 * * 0 := v A0 -> 1
00015 * * 1 L n 4
00016
00017 * * 4 L n 2 -> 5
00018
00019 * * 2 L v B1 -> 3
00020 * * 3 + c 1
00021
00022 * * 5 + v A1
00023
00024
00025 * 2 L v B1 -> 3
00026 * 3 + c 1
0010 D6 0D [3] 00027 ldab B1
0012 5C [2] 00028 incb
00029 * 4 L n 2 -> 5
00030 * 5 + v A1
0013 4F [2] 00031 clra
0014 DB 07 [3] 00032 addb A1+1
0016 99 06 [3] 00033 adca A1
00034 * 1 L n 4
00035 * 0 := v A0 -> 1
0018 D7 05 [4] 00036 stab A0+1
001A 97 04 [4] 00037 staa A0The generated comment lines show the original line of source code, the parse tree and the applicable nodes as code is generated.
Now the fun begins: playing with optimization of the tree.
How applicable this will be to compiling Python remains to be seen, but this is something I had been wanting to do for quite a long time.
From the previous post,
That was a test for the reader and nobody took the bait.
You may have noticed that the variable A0 was not allocated as “_A0” but as “_PTEST.A0”
For a simplistic language, PL/M has a sophisticated scoping mechanism. Procedures and functions may be defined. They may be nested, similar to what is allowed in Pascal. Furthermore, like in C, variables may be declared within a procedure or some nested DO blocks.
This is a valid PL/M program demonstrating nested scope:
P$Test: do;
declare A address; /* first A */
/* point 1 */
Proc: procedure;
declare A address; /* second A */
/* point 2 */
Block: do;
declare A address; /* third A */
/* point 3 */
end Block;
/* point 4 */
end Proc;
/* point 5 */
Another: do;
declare A address; /* fourth A */
/* point 6 */
end Another;
/* point 7 */
end P$Test;At points 1, 5 and 7, a reference to A gets the first A.
At points 2 and 4, a reference to A gets the second A.
At point 3, a reference to A gets the third A.
At point 6, a reference to A gets the fourth A.
The first A is allocated as _PTEST.A
The second as _PTEST.PROC.A
The third as _PTEST.PROC.BLOCK.A
And the fourth as _PTEST.ANOTHER.A
The compiler keeps track of block nesting to generate the name prefixes.
Scoping in Python is somewhat similiar to PL/M.
In this prior post,
[Project Log] Python on the 6502/C64, 8080, 6800, 6809 and AVR
an unexpected behavior of variable scoping in Python was discussed.
Building a parse tree is one way to uncover such forward “declarations” without having to make more than one pass of the source code.
Work continues on compiling Python, but do not be surprised to see a PL/M compiler for the 6502 appear as these techniques are developed further.
One way to subtract is to add the negative of a number.
PL/M has two special forms of add and subtract which takes into account the processor’s carry flag. That’s right, an add with carry or subtract with borrow in a “high-level” language.
Unfortunately, this is foiling the ability to save some code in the compiler by sharing the generation of addition and subtraction instructions. At least when generating code for the 6800.
ADC with the carry flag set adds an additional one; SBC subtracts an additional one. If a number is negated and added instead of subtracting, any borrow is not correctly handled, carrying (adding another 1) instead of borrowing (subtracting another 1.) If only the 6800 had an instruction to flip the carry flag like the 8080/Z80…
The 6502 may have gotten SBC right for this trick because it subtracts the inverse of the carry flag. The compiler can complement instead of negating the number for adding. That may be because the 6502 saved some transistors by complementing and adding instead of subtracting.
Let’s work through an example.
Suppose we want to subtract 1 from 0.
The low byte of 0 is %00000000; the high byte of 0 is %00000000.
The low byte of 1 is %00000001; the high byte of 1 is %00000000.
The low byte of -1 is %11111111; the high byte of -1 is %11111111.
Complementing is flipping all of the bits of a number (one’s complement)
Negating is flipping the bits then adding 1 (two’s complement)
The low byte of the one’s complement of 1 is %11111110; the high byte is %11111111.
If we subtract on the 6502,
0000 38 [2] 00001 sec ; No borrow
00002
0001 AD 0022 [4] 00003 lda Lo ; A contains %00000000
0004 E9 01 [2] 00004 sbc #1 ; A contains %11111111
0006 8D 0022 [4] 00005 sta Lo ; "carry" flag is clear
00006
0009 AD 0023 [4] 00007 lda Hi ; A contains %00000000
000C E9 00 [2] 00008 sbc #0 ; A contains %11111111
000E 8D 0023 [4] 00009 sta Hi ; "carry" flag is clearNote that the “carry” flag is clear signifying a borrow.
Now if we subtract by adding,
0011 38 [2] 00011 sec ; No borrow
00012
0012 AD 0022 [4] 00013 lda Lo ; A contains %00000000
0015 69 FE [2] 00014 adc #%11111110 ; A contains %11111111
0017 8D 0022 [4] 00015 sta Lo ; "carry" flag is clear
00016
001A AD 0023 [4] 00017 lda Hi ; A contains %00000000
001D 69 FF [2] 00018 adc #%11111111 ; A contains %11111111
001F 8D 0023 [4] 00019 sta Hi ; "carry" flag is clearThe results are identical.
If the initial sec is omitted, both code snippits accept a borrow from previous calculation.
However on the 6800,
0000 0C [2] 00001 clc ; No borrow
00002
0001 B6 0022 [4] 00003 ldaa Lo ; A contains %00000000
0004 82 01 [2] 00004 sbca #1 ; A contains %11111111
0006 B7 0022 [5] 00005 staa Lo ; "carry" flag is set
00006
0009 B6 0023 [4] 00007 ldaa Hi ; A contains %00000000
000C 82 00 [2] 00008 sbca #0 ; A contains %11111111
000E B7 0023 [5] 00009 staa Hi ; "carry" flag is setAnd by adding,
0011 0C [2] 00011 clc ; No borrow
00012
0012 B6 0022 [4] 00013 ldaa Lo ; A contains %00000000
0015 89 FE [2] 00014 adca #%11111110 ; A contains %11111110
0017 B7 0022 [5] 00015 staa Lo ; "carry" flag is clear
00016
001A B6 0023 [4] 00017 ldaa Hi ; A contains %00000000
001D 89 FF [2] 00018 adca #%11111111 ; A contains %11111111
001F B7 0023 [5] 00019 staa Hi ; "carry" flag is clearIf only there was an instruction to complement the carry flag, we can adopt the 6502 borrow convention and make the code work.
Still trying to climb out of this rabbit hole…
Web searching turned up some very interesting PL/M stuff.
The last discovery is particularly interesting. None of these use the PLUS and MINUS operators. These are the add with carry and subtract with borrow operations described previously.
Probably because even the PL/M-80 manual cautions about their use:
Trying to support MINUS definitely results in some fairly inefficient code on the 6800:
00016 **A0 = (A0 + A1) MINUS (A2 + A3);
00017
00018 * * 0 := v A0 -> 1
00019 * * 1 L n 4
00020
00021 * * 4 L n 2 -> 5
00022
00023 * * 2 L v A0 -> 3
00024 * * 3 + v A1
00025
00026 * * 5 -b n 6
00027
00028 * * 6 L v A2 -> 7
00029 * * 7 + v A3
00030
00031
00032
00033 * 2 L v A0 -> 3
00034 * 3 + v A1
0018 D6 05 [3] 00035 ldab A0+1
001A 96 04 [3] 00036 ldaa A0
001C DB 07 [3] 00037 addb A1+1
001E 99 06 [3] 00038 adca A1
00039 * 4 L n 2 -> 5
00040 * 5 -b n 6
0020 97 10 [4] 00041 staa Tp00_
0022 D7 11 [4] 00042 stab Tp00_+1
00043 * 6 L v A2 -> 7
00044 * 7 + v A3
0024 D6 09 [3] 00045 ldab A2+1
0026 96 08 [3] 00046 ldaa A2
0028 DB 0B [3] 00047 addb A3+1
002A 99 0A [3] 00048 adca A3
002C 97 12 [4] 00049 staa Tp01_
002E D7 13 [4] 00050 stab Tp01_+1
0030 96 10 [3] 00051 ldaa Tp00_
0032 D6 11 [3] 00052 ldab Tp00_+1
0034 D2 13 [3] 00053 sbcb Tp01_+1
0036 92 12 [3] 00054 sbca Tp01_
00055 * 1 L n 4
00056 * 0 := v A0 -> 1
0038 D7 05 [4] 00057 stab A0+1
003A 97 04 [4] 00058 staa A0Not being able to subtract by adding the complement necessitates using a second set of temporary variables on a machine with limited registers.
Code generation for addition and subtraction is complete. It has been refactored and is in good shape.
Working on code generation for and, or and xor is a joy in comparison because they
Part of the fun in code optimization is finding and exploiting opportunities in the way boolean logic works:
00016 **A0 = A1 and 0FF00h;
00017
00018 * * 0 := v A0 -> 1
00019 * * 1 L r 2
00020
00021 * * 2 L v A1 -> 3
00022 * * 3 a c -256
00023
00024
00025 * 2 L v A1 -> 3
00026 * 3 a c -256
0018 5F [2] 00027 clrb
0019 96 06 [3] 00028 ldaa A1
00029 * 1 L r 2
00030 * 0 := v A0 -> 1
001B D7 05 [4] 00031 stab A0+1
001D 97 04 [4] 00032 staa A0Edit: more fun with Boolean logic…
00016 **A0 = A1 or 0FFh;
00017
00018 * * 0 := v A0 -> 1
00019 * * 1 L r 2
00020
00021 * * 2 L v A1 -> 3
00022 * * 3 o c 255
00023
00024
00025 * 2 L v A1 -> 3
00026 * 3 o c 255
0018 C6 FF [2] 00027 ldab #$FF
001A 96 06 [3] 00028 ldaa A1
00029 * 1 L r 2
00030 * 0 := v A0 -> 1
001C D7 05 [4] 00031 stab A0+1
001E 97 04 [4] 00032 staa A0and
00016 **A0 = A1 + A2 xor 0FFh;
00017
00018 * * 0 := v A0 -> 1
00019 * * 1 L r 4
00020
00021 * * 4 L r 2 -> 5
00022
00023 * * 2 L v A1 -> 3
00024 * * 3 + v A2
00025
00026 * * 5 x c 255
00027
00028
00029 * 2 L v A1 -> 3
00030 * 3 + v A2
0018 D6 07 [3] 00031 ldab A1+1
001A 96 06 [3] 00032 ldaa A1
001C DB 09 [3] 00033 addb A2+1
001E 99 08 [3] 00034 adca A2
00035 * 4 L r 2 -> 5
00036 * 5 x c 255
0020 53 [2] 00037 comb
00038 * 1 L r 4
00039 * 0 := v A0 -> 1
0021 D7 05 [4] 00040 stab A0+1
0023 97 04 [4] 00041 staa A0PL/M defines true as a byte of all 1’s and false as a byte of all 0’s.
Implementing relational operators between two single-byte values is fairly easy.
00019 **B0 = B1 <> B2;
00020
00021 * * 0 := v B0 -> 1
00022 * * 1 L r 2
00023
00024 * * 2 L v B1 -> 3
00025 * * 3 NE v B2
00026
00027
00028 * 2 L v B1 -> 3
00029 * 3 NE v B2
0019 96 0E [3] 00030 ldaa B1
001B 5F [2] 00031 clrb
001C 90 0F [3] 00032 suba B2
001E 27 01 (0021)[4] 00033 beq L00000
0020 5A [2] 00034 decb
0021 00035 L00000
00036 * 1 L r 2
00037 * 0 := v B0 -> 1
0021 D7 0D [4] 00038 stab B0Likewise for comparing two address variables.
00019 **B0 = A1 = A2;
00020
00021 * * 0 := v B0 -> 1
00022 * * 1 L r 2
00023
00024 * * 2 L v A1 -> 3
00025 * * 3 EQ v A2
00026
00027
00028 * 2 L v A1 -> 3
00029 * 3 EQ v A2
0019 5F [2] 00030 clrb
001A DE 07 [4] 00031 ldx A1
001C 9C 09 [4] 00032 cpx A2
001E 26 01 (0021)[4] 00033 bne L00000
0020 5A [2] 00034 decb
0021 00035 L00000
00036 * 1 L r 2
00037 * 0 := v B0 -> 1
0021 D7 0D [4] 00038 stab B0Anything else gets quite involved.
00019 **B0 = A1 + A2 <> A3;
00020
00021 * * 0 := v B0 -> 1
00022 * * 1 L r 4
00023
00024 * * 4 L r 2 -> 5
00025
00026 * * 2 L v A1 -> 3
00027 * * 3 + v A2
00028
00029 * * 5 NE v A3
00030
00031
00032 * 2 L v A1 -> 3
00033 * 3 + v A2
0019 D6 08 [3] 00034 ldab A1+1
001B 96 07 [3] 00035 ldaa A1
001D DB 0A [3] 00036 addb A2+1
001F 99 09 [3] 00037 adca A2
00038 * 4 L r 2 -> 5
00039 * 5 NE v A3
0021 7F 0004 [6] 00040 clr Bool
0024 D0 0C [3] 00041 subb A3+1
0026 92 0B [3] 00042 sbca A3
0028 26 03 (002D)[4] 00043 bne L00001
002A 5D [2] 00044 tstb
002B 27 03 (0030)[4] 00045 beq L00000
002D 00046 L00001
002D 7A 0004 [6] 00047 dec Bool
0030 00048 L00000
0030 D6 04 [3] 00049 ldab Bool
00050 * 1 L r 4
00051 * 0 := v B0 -> 1
0032 D7 0D [4] 00052 stab B0This because the CPX instruction is “broken” on the 6800; only the Zero flag is set correctly.
Looking at some of the not-so-good code for relational comparisons made me wonder how well PL/M-80 did it.
I cannot get PL/M-80 for DOS to work. I may have to find an emulator of ISIS-II on an Intel MDS.
Study of the 8080 instruction set shows that is has limited 16-bit data handling capability. It can
That is it. No easy way to compare two 16-bit quantities without going through the single accumulator a byte at a time.
Well, the 16 bits, I would think would be mainly used in addressing rather than values.
16-bit numerical quantities would be needed to implement the CP/M filesystem.
Also, if the assembler was written in PL/M, there are many needs for 16-bit numbers.
A very interesting read about the early days of microprocessor development at Intel…
Still no luck finding a way to run the Intel PL/M 8080 compiler, but one of the manuals had some sample code in it.
Consider this PL/M procedure:
PRINT$STRING: PROCEDURE(NAME,LENGTH);
DECLARE NAME ADDRESS,
(LENGTH,I,CHAR BASED NAME) BYTE;
DO I = 0 to LENGTH-1;
CALL PRINT$CHAR(CHAR(I));
END;
END PRINT$STRING;Before we get to the code generated by the compiler, a few words about the compiler conventions.
Parameters to a procedure are allocated statically in memory unless the procedure is declared to be reentrant. Local variables are allocated next, contiguously.
If there are two or fewer parameters, they are passed in registers.
The first in register pair BC if ADDRESS and in register C if BYTE.
The second in register pair DE if ADDRESS and in register E if BYTE.
00C8 21 02E2 [10] 00078 lxi H,2E2h ; Point to NAME
00CB 71 [7] 00079 mov M,C ; Store NAME
00CC 23 [5] 00080 inx H
00CD 70 [7] 00081 mov M,B
00CE 2C [5] 00082 inr L ; Store LENGTH
00CF 73 [7] 00083 mov M,E
00D0 2C [5] 00084 inr L ; Clear I
00D1 36 00 [10] 00085 mvi M,0
00D3 21 02E4 [10] 00086 lxi H,2E4h ; Point to LENGTH
00D6 4E [7] 00087 mov C,M
00D7 0D [5] 00088 dcr C ; LENGTH-1
00D8 79 [5] 00089 mov A,C
00D9 2C [5] 00090 inr L
00DA 96 [7] 00091 sub M ; Compare with I
00DB DA 00F1 [10] 00092 jc 0F1h ; End loop
00DE 4E [7] 00093 mov C,M ; Make index into NAME
00DF 06 00 [7] 00094 mvi B,0
00E1 2A 02E2 [16] 00095 lhld 2E2h ; Add base of NAME
00E4 09 [10] 00096 dad B
00E5 7E [7] 00097 mov A,M ; Get next CHAR
00E6 4F [5] 00098 mov C,A
00E7 CD 00C0 [17] 00099 call 0C0h ; call PRINT$CHAR
00EA 21 02E5 [10] 00100 lxi H,2E5h ; Point to I
00ED 34 [10] 00101 inr M ; Increment I
00EE C3 00D3 [10] 00102 jmp 0D3h ; Back to top of the loop
00F1 C9 [10] 00103 retThis is not great code, but it is not bad. The compiler definitely knows about the strengths and weaknesses of the 8080.
Loading a byte from or storing a byte to an arbitrary location in memory takes three bytes of machine code and 13 clock cycles. And it has to go through the accumulator.
Contrast that with indirect access using an address in the HL register pair. The “memory” register M is an equal to the other single-byte registers except for slightly slower access.
This is what an assembly language programmer might write:
00C8 21 02E2 [10] 00107 lxi H,2E2h ; Point to NAME
00CB 71 [7] 00108 mov M,C ; Store NAME
00CC 23 [5] 00109 inx H
00CD 70 [7] 00110 mov M,B
00CE 23 [5] 00111 inx H ; Store LENGTH
00CF 73 [7] 00112 mov M,E
00D0 AF [4] 00113 xra A ; Check for LENGTH = 0
00D1 BB [4] 00114 cmp E
00D2 C8 [5/11] 00115 rz
00D3 2A 02E2 [16] 00116 Loop: lhld 2E2h ; Get next character
00D6 4E [7] 00117 mov C,M
00D7 23 [5] 00118 inx H ; Point to next character
00D8 22 02E2 [16] 00119 shld 2E2h ; Save for next time
00DB CD 00C0 [17] 00120 call 0C0h ; call PRINT$CHAR
00DE 21 02E4 [10] 00121 lxi H,2E4h ; Point to LENGTH
00E1 35 [10] 00122 dcr M ; Decrement LENGTH
00E2 C2 00D3 [10] 00123 jnz Loop
00E5 C9 [10] 00124 retNow I am curious what the compiler would generate if I had written PL/M code to do it this way…
Success! The Intel PL/M compiler lives!
Though I cannot figure out how to set the program origin. Not necessary for this test since I am not actually trying to build a working program.
The rewritten PL/M program:
PRINT$STRING: PROCEDURE(NAME,LENGTH);
DECLARE NAME ADDRESS,
(LENGTH,CHAR BASED NAME) BYTE;
DO WHILE LENGTH <> 0;
CALL PRINT$CHAR(CHAR);
NAME = NAME+1;
LENGTH = LENGTH-1;
END;
END PRINT$STRING;and the generated code:
000B 21 0004 [10] 00003 lxi H,0004
000E 73 [7] 00004 mov M,E
000F 2B [5] 00005 dcx H
0010 70 [7] 00006 mov M,B
0011 2B [5] 00007 dcx H
0012 71 [7] 00008 mov M,C
0013 3A 0004 [13] 00009 lda 0004
0016 FE 00 [7] 00010 cpi 00
0018 CA 001E [10] 00011 jz 0030
001B 2A 0002 [16] 00012 lhld 0002
001E 4E [7] 00013 mov C,M
001F CD 0000 [17] 00014 call 0000
0022 2A 0002 [16] 00015 lhld 0002
0025 23 [5] 00016 inx H
0026 22 0002 [16] 00017 shld 0002
0029 21 0004 [10] 00018 lxi H,0004
002C 35 [10] 00019 dcr M
002D C3 000D [10] 00020 jmp 0013
0030 C9 [10] 00021 retNot quite as good as hand-written code, but much improved.
Just when you thought you have seen everything, this test program came with the compiler. Cheers!
/*
* 99 bottles of beer in PL/M-80
*
* by John Durbetaki using AEDIT
*
*/
Ninety$Nine: do;
$include(:f1:common.lit)
$include(:f1:isis.ext)
declare as LITERALLY 'LITERALLY';
declare CRLF as '0Dh,0Ah';
declare Beers BYTE;
declare Message1(*) BYTE DATA(' of beer on the wall,',CRLF);
declare Message2(*) BYTE DATA(' of beeeeer . . . . ,',CRLF);
declare Message3(*) BYTE DATA('Take one down, pass it around,',CRLF);
declare Message4(*) BYTE DATA(' of beer on the wall.',CRLF);
declare End$Message(*) BYTE DATA(CRLF,'Time to buy more beer!',CRLF);
declare STATUS ADDRESS;
declare How$Many(128) BYTE;
declare How$Many$Count BYTE;
Copy: PROCEDURE(Ap,Bp,Count);
declare Ap ADDRESS;
declare A BASED Ap BYTE;
declare Bp ADDRESS;
declare B BASED Bp BYTE;
declare Count BYTE;
DO WHILE Count > 0;
B=A;
Ap=Ap+1;
Bp=Bp+1;
Count=Count-1;
END;
END;
Make$How$Many: PROCEDURE(Beers);
declare Beers BYTE;
if Beers = 0 THEN DO;
How$Many$Count=15;
CALL Copy(.('No more bottles'),.How$Many(0),How$Many$Count);
END;
else if Beers = 1 THEN DO;
How$Many$Count=15;
CALL Copy(.('One more bottle'),.How$Many(0),How$Many$Count);
END;
else DO;
if Beers >= 10 THEN DO;
How$Many(0)='0'+(Beers/10);
How$Many(1)='0'+(Beers MOD 10);
CALL Copy(.(' bottles'),.How$Many(2),8);
How$Many$Count=10;
END;
else DO;
How$Many(0)='0'+Beers;
CALL Copy(.(' bottles'),.How$Many(1),8);
How$Many$Count=9;
END;
END;
END;
Chug: PROCEDURE(Beers);
declare Beers BYTE;
CALL Make$How$Many(Beers);
CALL WRITE(0,.How$Many,How$Many$Count,.STATUS);
CALL WRITE(0,.Message1,SIZE(Message1),.STATUS);
CALL WRITE(0,.How$Many,How$Many$Count,.STATUS);
CALL WRITE(0,.Message2,SIZE(Message2),.STATUS);
CALL WRITE(0,.Message3,SIZE(Message3),.STATUS);
CALL Make$How$Many(Beers-1);
CALL WRITE(0,.How$Many,How$Many$Count,.STATUS);
CALL WRITE(0,.Message4,SIZE(Message4),.STATUS);
END;
Beers = /*99*/ 9;
DO WHILE Beers > 0;
CALL Chug(Beers);
Beers=Beers-1;
END;
CALL WRITE(0,.End$Message,SIZE(End$Message),.STATUS);
call exit;
END;Through the FLEX Users Group, I have discovered that Programma International sold SPL/M, a PL/M lookalike, running on and generating 6800 code for the FLEX operating system.