---
title: "Address-format code for Y=A*(B+D/C)+(G*E)/F"
subject: "Computer Organization and Architecture"
chapter: 2
key: "semester_5/computer_organization_and_architecture:address-code-3"
source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/address-code-3
site: BCT Engineering
---

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

Computer Organization and Architecture · Chapter 2: Central Processing Unit

- Asked in: 2073 Shrawan, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/address-code-3
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_organization_and_architecture/chapter-2

## Question

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

## Short answer

For Y = A*(B+D/C)+(G*E)/F, the counts are 6 three-address, 11 two-address, 10 one-address and 14 zero-address instructions. The postfix form A B D C / + * G E * F / + is used.

## Answer

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+(D/C)))+((G*E)/F)**. Postfix (reverse Polish) form: `A B D C / + * G E * F / +`.

#### Three-address instructions (OP dest, src1, src2)
```text
DIV R1, D, C   ; R1 = D / C
ADD R2, B, R1  ; R2 = B + R1
MUL R3, A, R2  ; R3 = A * R2
MUL R4, G, E   ; R4 = G * E
DIV R5, R4, F  ; R5 = R4 / F
ADD Y, R3, R5  ; Y = R3 + R5
```
Number of instructions: 6. Shortest program, but each instruction is long (three addresses).

#### Two-address instructions (OP dest, src; dest = dest OP src)
```text
MOV R1, A   ; R1 = A
MOV R2, B   ; R2 = B
MOV R3, D   ; R3 = D
DIV R3, C   ; R3 = R3 / C
ADD R2, R3  ; R2 = R2 + R3
MUL R1, R2  ; R1 = R1 * R2
MOV R2, G   ; R2 = G
MUL R2, E   ; R2 = R2 * E
DIV R2, F   ; R2 = R2 / F
ADD R1, R2  ; R1 = R1 + R2
MOV Y, R1   ; Y = R1
```
Number of instructions: 11. A MOV is needed first because the destination operand is overwritten.

#### One-address instructions (accumulator AC is the implied operand)
```text
LOAD D    ; AC = D
DIV C     ; AC = AC / C
ADD B     ; AC = AC + B
MUL A     ; AC = AC * A
STORE T1  ; T1 = AC
LOAD G    ; AC = G
MUL E     ; AC = AC * E
DIV F     ; AC = AC / F
ADD T1    ; AC = AC + T1
STORE Y   ; Y = AC
```
Number of instructions: 10. Intermediate results that cannot stay in AC are saved in memory temporaries.

#### Zero-address instructions (stack machine)
```text
PUSH A  ; push A
PUSH B  ; push B
PUSH D  ; push D
PUSH C  ; push C
DIV     ; pop two, compute, push result
ADD     ; pop two, compute, push result
MUL     ; pop two, compute, push result
PUSH G  ; push G
PUSH E  ; push E
MUL     ; pop two, compute, push result
PUSH F  ; push F
DIV     ; pop two, compute, push result
ADD     ; pop two, compute, push result
POP Y   ; Y = top of stack
```
Number of instructions: 14. 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 = 6, 2-address = 11, 1-address = 10, 0-address = 14 instructions. Fewer addresses per instruction means shorter instructions but more of them.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In Y = A*(B+D/C)+(G*E)/F, which sub-operation is computed first in the bracket B+D/C?
   A. B+D
   B. D/C
   C. A*B
   D. B+C
   Answer: B. Division has higher priority than addition, so D/C is done before adding B. (Key point: Order of evaluation)

2. What is the postfix (reverse Polish) form of Y = A*(B+D/C)+(G*E)/F?
   A. A B D C / + * G E * F / +
   B. A B D C + / * G E * F / +
   C. A B + D C / * G E * F / +
   D. A B D C / + * G E F * / +
   Answer: A. D C / is evaluated, added to B, multiplied by A; then G E * F / is formed and the two halves are added. (Key point: Postfix form)

3. How many three-address instructions does this expression need?
   A. 4
   B. 8
   C. 6
   D. 11
   Answer: C. Each of the six operations (/, +, *, *, /, +) takes one instruction, the last one storing into Y. (Key point: Three-address count)

4. In the zero-address (stack machine) program, which instructions contain a memory address?
   A. Every instruction
   B. Only ADD, SUB, MUL and DIV
   C. Only LOAD and STORE
   D. Only PUSH and POP
   Answer: D. Only PUSH and POP carry an address; arithmetic instructions work on the top two stack items. (Key point: Zero-address stack code)

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