Simple Computer Simulator Instruction-Set

--D. Thiebaut (talk) 16:57, 26 August 2014 (EDT)

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Instructions Using a Single Constant

These instructions operate with a single number (we refer to them as constants) that is either loaded into, added, or subtracted from the accumulator register.

Instruction	Code (decimal)	Code ( binary)	Description
ADD number	24	00011000	This instruction adds the number to the one already in the accumulator. For example, if the accumulator register already contains 10, and the processor executes ADD 3 the result is that the accumulator will contain 13 after the instruction.
COMP number	84	01010100	This instruction compares the number to the one already in the accumulator. For example, if the accumulator register already contains 10, and the processor executes COMP 3 the result is the comparison of 10 to 3. 10 is greater, and is not equal to 3. This will prevent a JLT (jump if less than) to jump to its target, and will prevent a JEQ instruction from jumping. On the other hand, if the accumulator had contained 2, then a COMP 2 would have allowed a subsequent JEQ instruction to jump to its target.
DIV number	44	00101100	This instruction divides the contents of the accumulator by number, and keeps the integer part of the result. For example, if the accumulator register already contains 10, and the processor executes DIV 4 the result is that the accumulator will contain 2 after the instruction.
LOAD number	4	00000100	This instruction puts the number into the accumulator. Whatever was in the accumulator prior to the operation is lost. , if the accumulator register already contains 10, and the processor executes ADD 3 the result is that the accumulator will contain 13 after the instruction.
MUL number	40	00101000	This instruction multiplies the contents of the accumulator by number, and replaces the contents of the accumulator by the result. For example, if the accumulator register already contains 10, and the processor executes MUL 4 the result is that the accumulator will contain 40 after the instruction.
SUB number	32	00100000	This instruction subtracts the number from the one already in the accumulator. For example, if the accumulator register already contains 10, and the processor executes SUB 3 the result is that the accumulator will contain 7 after the instruction.

Memory Instructions

Instruction	Code (decimal)	Code ( binary)	Description
ADDX	28	00011100
ADDXm	30	00011110
ADDXx	29	00011101
ADDm	26	00011010
ADDx	25	00011001
COMPX	92	01011100
COMPXm	94	01011110
COMPXx	93	01011101
COMPm	86	01010110
COMPx	85	01010101
DIVm	46	00101110
DIVx	45	00101101
HALT	127	01111111
JEQ	68	01000100
JLT	72	01001000
JUMP	64	01000000
LOADX	8	00001000
LOADXm	10	00001010
LOADXx	9	00001001
LOADm	6	00000110
LOADx	5	00000101
MULm	42	00101010
MULx	41	00101001
NOP	0	00000000
STOREXm	22	00010110
STOREXx	21	00010101
STOREm	18	00010010
STOREx	17	00010001
SUBX	36	00100100
SUBXm	38	00100110
SUBXx	37	00100101
SUBm	34	00100010
SUBx	33	00100001
TAX	79	01001111
TXA	83	01010011