---
title: "Address-format code for Y=A/B+(C×D)+F(H/G)"
subject: "Computer Organization and Architecture"
chapter: 2
key: "semester_5/computer_organization_and_architecture:address-code-8"
source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/address-code-8
site: BCT Engineering
---

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

Computer Organization and Architecture · Chapter 2: Central Processing Unit

- Asked in: 2070 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-8
- 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+(C×D)+F(H/G).

## Short answer

For Y = A/B+(C×D)+F(H/G), the postfix form is A B / C D * + F H G / * +. Three-address code uses 6 instructions (DIV, MUL, ADD, DIV, MUL, ADD). Two-address requires 11 instructions with MOV overhead. One-address (accumulator) needs 12 instructions using temporaries T1, T2. Zero-address (stack) uses 14 instructions following postfix directly with PUSH/POP. Fewer addresses per instruction means shorter instructions but more total instructions.

## 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. Adjacent letters with no operator, or a letter before a bracket, are read as multiplication. So F(H/G) is F × (H/G).

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

#### Three-address instructions (OP dest, src1, src2)
```text
DIV R1, A, B    ; R1 = A / B
MUL R2, C, D    ; R2 = C * D
ADD R3, R1, R2  ; R3 = R1 + R2
DIV R4, H, G    ; R4 = H / G
MUL R5, F, R4   ; R5 = F * R4
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
DIV R1, B   ; R1 = R1 / B
MOV R2, C   ; R2 = C
MUL R2, D   ; R2 = R2 * D
ADD R1, R2  ; R1 = R1 + R2
MOV R2, F   ; R2 = F
MOV R3, H   ; R3 = H
DIV R3, G   ; R3 = R3 / G
MUL R2, R3  ; R2 = R2 * R3
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 A    ; AC = A
DIV B     ; AC = AC / B
STORE T1  ; T1 = AC
LOAD C    ; AC = C
MUL D     ; AC = AC * D
ADD T1    ; AC = AC + T1
STORE T2  ; T2 = AC
LOAD H    ; AC = H
DIV G     ; AC = AC / G
MUL F     ; AC = AC * F
ADD T2    ; AC = AC + T2
STORE Y   ; Y = AC
```
Number of instructions: 12. 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
DIV     ; pop two, compute, push result
PUSH C  ; push C
PUSH D  ; push D
MUL     ; pop two, compute, push result
ADD     ; pop two, compute, push result
PUSH F  ; push F
PUSH H  ; push H
PUSH G  ; push G
DIV     ; pop two, compute, push result
MUL     ; 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 = 12, 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. How is F(H/G) in Y = A/B+(C×D)+F(H/G) interpreted?
   A. As F added to H/G
   B. As the function F applied to H/G
   C. As F divided by (H/G)
   D. As F multiplied by (H/G)
   Answer: D. A letter before a bracket with no operator is read as multiplication. (Key point: Reading F(H/G))

2. What is the postfix form of the expression?
   A. A B / C D * + F H G / * +
   B. A B / C D * + F H / G * +
   C. A B / C D + * F H G / * +
   D. A B / C D * + F H G * / +
   Answer: A. A/B and C*D are added; H G / is multiplied by F; the two parts are added. (Key point: Postfix form)

3. How many three-address instructions are required?
   A. 4
   B. 6
   C. 9
   D. 12
   Answer: B. Six operators (/, *, +, /, *, +) give six instructions. (Key point: Three-address count)

4. In one-address code, why are STORE Tn instructions needed in the middle of the program?
   A. To convert the expression to postfix form
   B. Because the ALU cannot add numbers directly
   C. The accumulator is the only implied register, so an intermediate result must be saved in memory before AC is reused
   D. To increase the number of addresses per instruction
   Answer: C. AC holds only one value; a partial result needed later has to be stored in a temporary memory location. (Key point: Temporaries in one-address code)

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