--- 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-16" source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/address-code-16 site: BCT Engineering --- # Address-format code for Y=A+(B*C)+D Computer Organization and Architecture · Chapter 2: Central Processing Unit - Asked in: 2068 Baisakh (old course), 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-16 - Whole chapter: https://www.bctengineer.com/md/semester_5/computer_organization_and_architecture/chapter-2 ## Question Write down the following equation in three addresses, two addresses, one address and zero address instruction. If necessary, use temporary location T to store intermediate result. Y = A + (B*C) + D. ## Short answer Y = A+(B*C)+D requires three three-address, six two-address, five one-address, and eight zero-address instructions. The one-address code needs no temporaries because the accumulator can hold all intermediate results. ## 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. Left-to-right evaluation gives (A + (B×C)) + D. Fully bracketed: **Y = (A+(B*C))+D**. Postfix (reverse Polish) form: `A B C * + D +`. #### Three-address instructions (OP dest, src1, src2) ```text MUL R1, B, C ; R1 = B * C ADD R2, A, R1 ; R2 = A + R1 ADD Y, R2, D ; Y = R2 + D ``` 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 MOV R2, B ; R2 = B MUL R2, C ; R2 = R2 * C ADD R1, R2 ; R1 = R1 + R2 ADD R1, D ; R1 = R1 + D 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 B ; AC = B MUL C ; AC = AC * C ADD A ; AC = AC + A ADD D ; AC = AC + D STORE Y ; Y = AC ``` Number of instructions: 5. No temporary location T is needed here, because every intermediate result can stay in AC (addition is commutative, so B×C is computed first and A is added to it). #### Zero-address instructions (stack machine) ```text PUSH A ; push A PUSH B ; push B PUSH C ; push C MUL ; pop two, compute, push result ADD ; pop two, compute, push result PUSH D ; push D ADD ; 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 = 5, 0-address = 8 instructions. Fewer addresses per instruction usually 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 Y = A + (B*C) + D evaluated? A. As A + ((B*C) + D) using multiplication last B. As (A + (B*C)) + D C. As (A + B) * (C + D) D. As A + B * (C + D) Answer: B. Equal-priority additions go left to right, with B*C computed first because of the bracket. (Key point: Left-to-right rule) 2. What is the postfix form? A. A B C * + D + B. A B + C * D + C. A B C + * D + D. A B C * D + + Answer: A. B C * is added to A, then D is added. (Key point: Postfix form) 3. Why does the one-address code need no STORE to a temporary here? A. The accumulator holds two values at once B. Y is used as a temporary C. B*C can be computed in AC, then A and D are added to AC directly D. Multiplication results are stored automatically Answer: C. All partial results stay in AC because each operation uses the previous result, so only the final STORE Y is needed. (Key point: One-address advantage) 4. Which is the correct set of instruction counts (3-, 2-, 1-, 0-address)? A. 3, 5, 6, 8 B. 6, 3, 5, 8 C. 3, 6, 8, 5 D. 3, 6, 5, 8 Answer: D. The counts are 3, 6, 5 and 8 for three-, two-, one- and zero-address code. 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