- Asked 16/22
- 2072 Kartik · 8 marks
Address-format code for 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.
Write codes using 3, 2, 1 and 0 address instruction formats to perform the operation Y = (A+B)(C+D)+G/EF.
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)
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)
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)
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)
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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