---
title: "Address-format code for Y=(A+B)/C+D/(E*F)"
subject: "Computer Organization and Architecture"
chapter: 2
key: "semester_5/computer_organization_and_architecture:address-code-14"
source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/address-code-14
site: BCT Engineering
---

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

Computer Organization and Architecture · Chapter 2: Central Processing Unit

- Asked in: 2075 Chaitra, 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-14
- 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/(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.

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

#### Three-address instructions (OP dest, src1, src2)
```text
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)
```text
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)
```text
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)
```text
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.

## Check yourself

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

1. In Y = (A+B)/C + D/(E*F), what must be computed before D can be divided?
   A. E*F
   B. A+B
   C. C+D
   D. D+E
   Answer: A. The bracket (E*F) is evaluated first, then D is divided by it. (Key point: Order of evaluation)

2. What is the postfix form of the expression?
   A. A B + C / D E * F / +
   B. A B + C / D E F * / +
   C. A B C + / D E F * / +
   D. A B + C / D E F / * +
   Answer: B. A B + C / gives the first term; E F * then D ... / gives the second; + joins them. (Key point: Postfix form)

3. How many three-address instructions are required?
   A. 4
   B. 6
   C. 5
   D. 9
   Answer: C. Five operators (+, /, *, /, +) give five instructions. (Key point: Three-address count)

4. In two-address code (OP dest, src), why is a MOV usually needed before an arithmetic instruction?
   A. Memory cannot be used as a source operand
   B. The destination operand is overwritten by the result, so the first value must be copied into a register first
   C. MOV makes the instruction shorter
   D. The ALU can only add values that were moved from memory
   Answer: B. In dest = dest OP src the destination is overwritten, so the original variable is copied into a register with MOV to keep it intact. (Key point: Why MOV in two-address code)

## Close variants set in other exams

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- [Address-format code for X=(A-B*F)*C+D/E](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Aaddress-code-6) — Computer Organization and Architecture, chapter 2, asked 16×
- [Address-format code for Y=A/B+(C×D)+F(H/G)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Aaddress-code-8) — Computer Organization and Architecture, chapter 2, asked 16×
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- [Address-format code for Y=(A-B/C)×(D+E×G)/F](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Aaddress-code-11) — Computer Organization and Architecture, chapter 2, asked 16×
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- [Address-format code for Y=AB+(F/G)+CD](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Aaddress-code-17) — Computer Organization and Architecture, chapter 2, asked 16×

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