multiply_two_numbers.md

Program: Multiply two numbers

Example program that calculates 7x8 and displays the result on the display.

#	Instruction	Address	Memory	Comment
0	LDA 14	0000	0001 1110	Put the value of the X variable in the A-register
1	SUB 12	0001	0011 1100	Put the value of the C constant in the B-register and store A-B in the A-register
2	JC 6	0010	0111 0110	Jump to instruction 6 if the A-register is past 255
3	LDA 13	0011	0001 1101	Put the value of the P variable in the A-register
4	OUT	0100	1110 0000	Output the value of the A-register
5	HLT	0101	1111 0000	Halt the computer
6	STA 14	0110	0100 1110	Store the value from the A-register in the X variable
7	LDA 13	0111	0001 1101	Put the value of the P variable in the A-register
8	ADD 15	1000	0010 1111	Put the value of the Y variable in the B-register and store A+B in the A-register
9	STA 13	1001	0100 1101	Store the value from the A-register in the P variable
10	JMP 0	1010	0110 0000	Jump to instruction 0
12	C	1100	0000 0001	The C constant predefined with the value 1
13	P	1101	0000 0000	The P (product/answer) variable predefined with the value 0
14	X	1110	0000 0111	The X (multiplicand) variable predefined with the value 7
15	Y	1111	0000 1000	The Y (multiplier) variable predefined with the value 8

The computer does not have a dedicated instruction for multiplication, so the program adds in a loop. The algorithm can be translated to this C code:

int c = 1, p = 0, x = 7, y = 8;
while (x > 0) {
    x = x - c;
    p = p + y;
}
printf("%d", p);

The assembly doesn't immediately make sense, as it looks like it would skip the jump with carry and output 0 and halt, due to the first subtraction resulting in the value 6 and not something past 255. The trick is the usage of two's complement when subtracting in the ALU. The subtract instruction tells the ALU to treat the value 1 from the B-register as -1 and add the numbers. -1 is 1111 1111 in two's compliment. So the calculation turns into 0000 0111 (7) + 1111 1111 (255) which is 1 0000 0110 (262), or 0000 0110 (6) with carry since we only have 8 bits and a carry. That jump with carry is actually very useful in this case, as it can be used to easily decrement X by 1 until it's 0, in which case 0 + 255 is 255 and no more carry so the loop ends and the result is printed.

The subtractions are calculated as follows:

7 - 1	->	7 + -1 (255)	=	0000 0111 + 1111 1111	=	1 0000 0110	=	262	=	6+C
6 - 1	->	6 + -1 (255)	=	0000 0110 + 1111 1111	=	1 0000 0101	=	261	=	5+C
5 - 1	->	5 + -1 (255)	=	0000 0101 + 1111 1111	=	1 0000 0100	=	260	=	4+C
4 - 1	->	4 + -1 (255)	=	0000 0100 + 1111 1111	=	1 0000 0011	=	259	=	3+C
3 - 1	->	3 + -1 (255)	=	0000 0011 + 1111 1111	=	1 0000 0010	=	258	=	2+C
2 - 1	->	2 + -1 (255)	=	0000 0010 + 1111 1111	=	1 0000 0001	=	257	=	1+C
1 - 1	->	1 + -1 (255)	=	0000 0001 + 1111 1111	=	1 0000 0000	=	256	=	0+C
0 - 1	->	0 + -1 (255)	=	0000 0000 + 1111 1111	=	1111 1111	=	255	=	255

This table displays the values of registers, variables and output during all iterations of the program.

	0	1	2	3	4	5	6	7
A	8	16	24	32	40	48	56	56
B	8	8	8	8	8	8	8	1
C	1	1	1	1	1	1	1	1
P	8	16	24	32	40	48	56	56
X	6	5	4	3	2	1	0	0
Y	8	8	8	8	8	8	8	8
OUT	x	x	x	x	x	x	x	56

This program is from Conditional jump instructions.