- Asked 16/22
- 2067 Asar (old course) · 8 marks
Address-format code for Y=AB+(F/G)+CD
Short answer
The postfix expression is A B * F G / + C D * +. Three-address code uses 5 instructions, two-address 9, one-address 11, zero-address 12. Fewer addresses per instruction results in more instructions overall.
Write codes using 3, 2, 1 and 0 address instruction formats to perform the operation Y = AB+(F/G)+CD.
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. Adjacent letters with no operator, or a letter before a bracket, are read as multiplication. So AB is A × B and CD is C × D.
Fully bracketed: Y = ((AB)+(F/G))+(CD). Postfix (reverse Polish) form: A B * F G / + C D * +.
Three-address instructions (OP dest, src1, src2)
MUL R1, A, B ; R1 = A * B
DIV R2, F, G ; R2 = F / G
ADD R3, R1, R2 ; R3 = R1 + R2
MUL R4, C, D ; R4 = C * D
ADD Y, R3, R4 ; Y = R3 + R4
Number of instructions: 5. Shortest program, but each instruction is long (three addresses).
Two-address instructions (OP dest, src; dest = dest OP src)
MOV R1, A ; R1 = A
MUL R1, B ; R1 = R1 * B
MOV R2, F ; R2 = F
DIV R2, G ; R2 = R2 / G
ADD R1, R2 ; R1 = R1 + R2
MOV R2, C ; R2 = C
MUL R2, D ; R2 = R2 * D
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)
LOAD A ; AC = A
MUL B ; AC = AC * B
STORE T1 ; T1 = AC
LOAD F ; AC = F
DIV G ; AC = AC / G
ADD T1 ; AC = AC + T1
STORE T2 ; T2 = AC
LOAD C ; AC = C
MUL D ; AC = AC * D
ADD T2 ; AC = AC + T2
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)
PUSH A ; push A
PUSH B ; push B
MUL ; pop two, compute, push result
PUSH F ; push F
PUSH G ; push G
DIV ; pop two, compute, push result
ADD ; pop two, compute, push result
PUSH C ; push C
PUSH D ; push D
MUL ; 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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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