--- title: "Address-format code for Y=(A+B)*(C+D)" subject: "Computer Organization and Architecture" chapter: 2 key: "semester_5/computer_organization_and_architecture:address-code-7" source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/address-code-7 site: BCT Engineering --- # Address-format code for Y=(A+B)*(C+D) Computer Organization and Architecture · Chapter 2: Central Processing Unit - Asked in: 2071 Chaitra, 5 marks; 2071 Shrawan, 6 marks (asked 2 times) - 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-7 - 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). ## Short answer Using postfix A B + C D + *. Three-address: 3 instructions (ADD R1,A,B; ADD R2,C,D; MUL Y,R1,R2). Two-address: 6 (MOV R1,A; ADD R1,B; MOV R2,C; ADD R2,D; MUL R1,R2; MOV Y,R1). One-address: 7 (LOAD A; ADD B; STORE T1; LOAD C; ADD D; MUL T1; STORE Y). Zero-address: 8 (PUSH A; PUSH B; ADD; PUSH C; PUSH D; ADD; MUL; POP Y). ## 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)**. Postfix (reverse Polish) form: `A B + C D + *`. #### Three-address instructions (OP dest, src1, src2) ```text ADD R1, A, B ; R1 = A + B ADD R2, C, D ; R2 = C + D MUL Y, R1, R2 ; Y = R1 * R2 ``` Number of instructions: 3. 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 Y, R1 ; Y = R1 ``` Number of instructions: 6. 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 Y ; Y = AC ``` Number of instructions: 7. 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 POP Y ; Y = top of stack ``` Number of instructions: 8. 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 = 3, 2-address = 6, 1-address = 7, 0-address = 8 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. The postfix form of (A+B)*(C+D) is A. A B C D + + * B. A B + C D + * C. A B * C D * + D. + A B + C D * Answer: B. Operators follow operands. (Key point: Postfix form) 2. How many instructions does the two-address code need? A. 3 B. 7 C. 8 D. 6 Answer: D. MOV, ADD, MOV, ADD, MUL, MOV. (Key point: Two-address count) 3. Why is a MOV needed first in two-address code? A. To clear the status flags B. The destination operand is overwritten C. To set the program counter D. To reduce the instruction count Answer: B. dest = dest OP src. (Key point: Why MOV first) 4. How many instructions does zero-address code need? A. 3 B. 6 C. 7 D. 8 Answer: D. Four PUSH, two ADD, one MUL and one POP make 8 instructions. 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