---
title: "Address-format code for X=(A-B*F)*C+D/E"
subject: "Computer Organization and Architecture"
chapter: 2
key: "semester_5/computer_organization_and_architecture:address-code-6"
source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/address-code-6
site: BCT Engineering
---

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

Computer Organization and Architecture · Chapter 2: Central Processing Unit

- Asked in: 2072 Chaitra, 8 marks
- Video: [Instruction Format | zero, one, two and three address instructions program | CAO | COA](https://www.youtube.com/watch?v=Htw_NTvj1ss) — Techno Tutorials ( e-Learning)
- 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-6
- 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 X = (A-B*F)*C+D/E.

## Short answer

Using postfix A B F * - C * D E / +, the operation needs 5 three-address, 9 two-address, 11 one-address and 12 zero-address instructions.

## Answer

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

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

#### Three-address instructions (OP dest, src1, src2)
```text
MUL R1, B, F   ; R1 = B * F
SUB R2, A, R1  ; R2 = A - R1
MUL R3, R2, C  ; R3 = R2 * C
DIV R4, D, E   ; R4 = D / E
ADD X, R3, R4  ; X = R3 + R4
```
Number of instructions: 5. 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
MUL R2, F   ; R2 = R2 * F
SUB R1, R2  ; R1 = R1 - R2
MUL R1, C   ; R1 = R1 * C
MOV R2, D   ; R2 = D
DIV R2, E   ; R2 = R2 / E
ADD R1, R2  ; R1 = R1 + R2
MOV X, R1   ; X = 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)
```text
LOAD B    ; AC = B
MUL F     ; AC = AC * F
STORE T1  ; T1 = AC
LOAD A    ; AC = A
SUB T1    ; AC = AC - T1
MUL C     ; AC = AC * C
STORE T2  ; T2 = AC
LOAD D    ; AC = D
DIV E     ; AC = AC / E
ADD T2    ; AC = AC + T2
STORE X   ; X = AC
```
Number of instructions: 11. 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 F  ; push F
MUL     ; pop two, compute, push result
SUB     ; pop two, compute, push result
PUSH C  ; push C
MUL     ; pop two, compute, push result
PUSH D  ; push D
PUSH E  ; push E
DIV     ; pop two, compute, push result
ADD     ; pop two, compute, push result
POP X   ; X = 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.

## Check yourself

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

1. In X = (A-B*F)*C+D/E, what is done first inside the bracket?
   A. B*F
   B. A-B
   C. A-F
   D. C+D
   Answer: A. Multiplication has higher priority than subtraction, so B*F is computed before subtracting from A. (Key point: Order of evaluation)

2. What is the postfix form of X = (A-B*F)*C+D/E?
   A. A B - F * C * D E / +
   B. A B F * - C * D E / +
   C. A B F - * C * D E / +
   D. A B F * - C D E / * +
   Answer: B. B F * is subtracted from A, multiplied by C, then D E / is added. (Key point: Postfix form)

3. How many three-address instructions are required?
   A. 4
   B. 7
   C. 5
   D. 9
   Answer: C. There are five operators (*, -, *, /, +), one instruction each. (Key point: Three-address count)

4. In the zero-address (stack machine) program, which instructions contain a memory address?
   A. Every instruction
   B. Only PUSH and POP
   C. Only ADD, SUB, MUL and DIV
   D. Only LOAD and STORE
   Answer: B. 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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