---
title: "Address-format code for Y=(A+B)*(C+D)+G/E*F"
subject: "Computer Organization and Architecture"
chapter: 2
key: "semester_5/computer_organization_and_architecture:address-code-5"
source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/address-code-5
site: BCT Engineering
---

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

Computer Organization and Architecture · Chapter 2: Central Processing Unit

- Asked in: 2072 Kartik, 8 marks
- Video: [Instruction Format | zero, one, two and three address instructions program | COA by Pro Academy MCA](https://www.youtube.com/watch?v=KVN7WQqTSzI) — Pro Academy 👍
- 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-5
- 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)+G/E*F.

## Short answer

For Y = (A+B)*(C+D)+G/E*F, postfix A B + C D + * G E / F * + gives 6 three-address, 10 two-address, 12 one-address and 14 zero-address 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. G/E*F is evaluated left to right as (G/E)*F.

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

#### Three-address instructions (OP dest, src1, src2)
```text
ADD R1, A, B    ; R1 = A + B
ADD R2, C, D    ; R2 = C + D
MUL R3, R1, R2  ; R3 = R1 * R2
DIV R4, G, E    ; R4 = G / E
MUL 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
ADD R1, B   ; R1 = R1 + B
MOV R2, C   ; R2 = C
ADD R2, D   ; R2 = R2 + D
MUL R1, R2  ; R1 = R1 * R2
MOV R2, G   ; R2 = G
DIV R2, E   ; R2 = R2 / E
MUL R2, F   ; R2 = R2 * F
ADD R1, R2  ; R1 = R1 + R2
MOV Y, R1   ; Y = R1
```
Number of instructions: 10. 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
ADD B     ; AC = AC + B
STORE T1  ; T1 = AC
LOAD C    ; AC = C
ADD D     ; AC = AC + D
MUL T1    ; AC = AC * T1
STORE T2  ; T2 = AC
LOAD G    ; AC = G
DIV E     ; AC = AC / E
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
ADD     ; pop two, compute, push result
PUSH C  ; push C
PUSH D  ; push D
ADD     ; pop two, compute, push result
MUL     ; pop two, compute, push result
PUSH G  ; push G
PUSH E  ; push E
DIV     ; pop two, compute, push result
PUSH F  ; push F
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 = 10, 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 G/E*F evaluated in Y = (A+B)*(C+D)+G/E*F?
   A. As G/(E*F)
   B. As (G/E)*F, because equal-priority operators go left to right
   C. As G/E first then added to F
   D. As (G*F)/E only if F is in a register
   Answer: B. Division and multiplication have equal priority, so they are done from left to right. (Key point: Left-to-right rule)

2. What is the postfix form of the expression?
   A. A B + C D + * G E F * / +
   B. A B C D + + * G E / F * +
   C. A B + C D + * G E / F * +
   D. A B + C D + G E / F * * +
   Answer: C. (A+B)(C+D) gives A B + C D + *, and (G/E)*F gives G E / F *, joined by +. (Key point: Postfix form)

3. What is the trade-off as the number of addresses per instruction decreases (3 to 0)?
   A. Instructions become shorter but more of them are needed
   B. Instructions become longer and fewer
   C. The program size stays the same
   D. Fewer instructions and shorter instructions both
   Answer: A. Fewer address fields make each instruction short, but more instructions (MOV, LOAD, STORE, PUSH) are needed. (Key point: Trade-off of address count)

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