Skip to main content

Address-format code for Y=(A+B)/C+D/(E*F)

Short answer

Y = (A+B)/C + D/(E*F) takes five three-address, nine two-address, eleven one-address, and twelve zero-address instructions. The one-address program stores partial results in memory temporaries.

Write codes using 3, 2, 1 and 0 address instruction formats to perform the operation Y = (A+B)/C + D/(E*F).

Assumptions: the operands are memory locations, R1, R2, ... are CPU registers, T1, T2, ... are temporary memory locations, and Y receives the result. Multiplication and division are done before addition and subtraction, and equal-priority operators go left to right.

Fully bracketed: Y = ((A+B)/C)+(D/(E*F)). Postfix (reverse Polish) form: A B + C / D E F * / +.

Three-address instructions (OP dest, src1, src2)

ADD R1, A, B   ; R1 = A + B
DIV R2, R1, C  ; R2 = R1 / C
MUL R3, E, F   ; R3 = E * F
DIV R4, D, R3  ; R4 = D / R3
ADD Y, R2, R4  ; Y = R2 + R4

Number of instructions: 5. Shortest program, but each instruction is long (three addresses).

Two-address instructions (OP dest, src; dest = dest OP src)

MOV R1, A   ; R1 = A
ADD R1, B   ; R1 = R1 + B
DIV R1, C   ; R1 = R1 / C
MOV R2, D   ; R2 = D
MOV R3, E   ; R3 = E
MUL R3, F   ; R3 = R3 * F
DIV R2, R3  ; R2 = R2 / R3
ADD R1, R2  ; R1 = R1 + R2
MOV Y, R1   ; Y = R1

Number of instructions: 9. A MOV is needed first because the destination operand is overwritten.

One-address instructions (accumulator AC is the implied operand)

LOAD A    ; AC = A
ADD B     ; AC = AC + B
DIV C     ; AC = AC / C
STORE T1  ; T1 = AC
LOAD E    ; AC = E
MUL F     ; AC = AC * F
STORE T2  ; T2 = AC
LOAD D    ; AC = D
DIV T2    ; AC = AC / T2
ADD T1    ; AC = AC + T1
STORE Y   ; Y = AC

Number of instructions: 11. Intermediate results that cannot stay in AC are saved in memory temporaries.

Zero-address instructions (stack machine)

PUSH A  ; push A
PUSH B  ; push B
ADD     ; pop two, compute, push result
PUSH C  ; push C
DIV     ; pop two, compute, push result
PUSH D  ; push D
PUSH E  ; push E
PUSH F  ; push F
MUL     ; pop two, compute, push result
DIV     ; pop two, compute, push result
ADD     ; pop two, compute, push result
POP Y   ; Y = top of stack

Number of instructions: 12. Only PUSH and POP carry an address; ADD, SUB, MUL, DIV work on the top two stack items, so the program follows the postfix form directly.

Summary: 3-address = 5, 2-address = 9, 1-address = 11, 0-address = 12 instructions. Fewer addresses per instruction means shorter instructions but more of them.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Computer Organization and Architecture chapter 2 questions →