--- title: "Address-format code for X=(A*B/C)-(D+E/F)" subject: "Computer Organization and Architecture" chapter: 2 key: "semester_5/computer_organization_and_architecture:address-code-1" source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/address-code-1 site: BCT Engineering --- # Address-format code for X=(A*B/C)-(D+E/F) Computer Organization and Architecture · Chapter 2: Central Processing Unit - Asked in: 2074 Asoj, 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-1 - 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/C)-(D+E/F). ## Short answer Using postfix form A B * C / D E F / + -, the operation needs 5 three-address, 9 two-address, 9 one-address (accumulator) and 12 zero-address instructions. Fewer addresses per instruction reduce instruction length but increase count. ## 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)/C)-(D+(E/F))**. Postfix (reverse Polish) form: `A B * C / D E F / + -`. #### Three-address instructions (OP dest, src1, src2) ```text MUL R1, A, B ; R1 = A * B DIV R2, R1, C ; R2 = R1 / C DIV R3, E, F ; R3 = E / F ADD R4, D, R3 ; R4 = D + R3 SUB X, R2, R4 ; X = 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 MUL R1, B ; R1 = R1 * B DIV R1, C ; R1 = R1 / C MOV R2, D ; R2 = D MOV R3, E ; R3 = E DIV R3, F ; R3 = R3 / F ADD R2, R3 ; R2 = R2 + R3 SUB 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 E ; AC = E DIV F ; AC = AC / F ADD D ; AC = AC + D STORE T1 ; T1 = AC LOAD A ; AC = A MUL B ; AC = AC * B DIV C ; AC = AC / C SUB T1 ; AC = AC - T1 STORE X ; X = AC ``` Number of instructions: 9. 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 MUL ; 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 DIV ; pop two, compute, push result ADD ; pop two, compute, push result SUB ; 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 = 9, 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. Postfix of (A*B/C)-(D+E/F) is 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. Follows precedence. (Key point: Postfix) 2. The three-address program has how many instructions? A. 3 B. 9 C. 12 D. 5 Answer: D. MUL, DIV, DIV, ADD, SUB. (Key point: Three-address count) 3. The zero-address program has how many instructions? A. 9 B. 12 C. 5 D. 8 Answer: B. Six PUSH, five operations (MUL, DIV, DIV, ADD, SUB) and one POP make 12. (Key point: Zero-address count) 4. Why is E/F computed before adding D? A. Addition is always done before division B. Subtraction must be done first C. Operands must be in alphabetical order D. Division has higher priority than addition Answer: D. Division is done before addition. 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