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Ingy döt Net 2023-07-01 11:58:00 -04:00
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PROGRAM SUBLEQ0 !Simulates a One-Instruction computer, with Subtract and Branch if <= 0.
INTEGER LOTS,LOAD !Document some bounds.
PARAMETER (LOTS = 36, LOAD = 31) !Sufficient for the example.
INTEGER IAR, MEM(0:LOTS) !The basic storage of a computer. IAR could be in memory too.
INTEGER ABC(3),A,B,C !A hardware register. Could use INTEGER*1 for everything...
EQUIVALENCE (ABC(1),A),(ABC(2),B),(ABC(3),C) !It has components.
INTEGER INITIAL(0:LOAD) !There is no sign of a bootstrap loader sequence!
DATA INITIAL/15,17,-1,17,-1,-1,16,1,-1,16,3,-1,15,15,0,0,-1, !These are operations, it so happens.
1 72,101,108,108,111,44,32,119,111,114,108,100,33,10,0/ !And these happen to be ASCII character code numbers.
Core memory initialisation.
MEM = -66 !Accessing uninitialised memory is improper. This might cause hiccoughs..
MEM(0:LOAD) = INITIAL !No bootstrap!
IAR = 0 !The Instruction Address Register starts at the start.
Commence execution of the current instruction.
100 ABC = MEM(IAR:IAR + 2) !Load the three-word instruction.
IAR = IAR + 3 !Advance IAR accordingly.
IF (A .EQ. -1) THEN !Decode the instruction as per the design.
WRITE (6,102) !Supply a prompt, otherwise, obscurity results.
102 FORMAT (" A number:",$) !But, that will make a mess of the layout.
READ (5,*) MEM(B) !The specified action is to read as a number.
ELSE IF (B .EQ. -1) THEN !This is for output.
WRITE (6,103) CHAR(MEM(A)) !As specified, interpret a number as a character.
103 FORMAT (A1,$) !The $, obviously, states: do not end the line and start the next.
ELSE !And this is a two-part action.
MEM(B) = MEM(B) - MEM(A) !Perform arithmetic.
IF (MEM(B).LE.0) IAR = C !And based on the result, maybe a GO TO.
END IF !So much for decoding.
IF (IAR.GE.0) GO TO 100 !Keep at it.
END !That was simple.