Question

Using the generic three-, two-, one-, and zero-address instruction sets (like in the notes), create programs to calculate (A+3)*B+C*5. You can assume that memory-memory operations are allowed. You may also define temporary locations if needed.

Answer 1

zero address instrutions

PUSH A //Top of Stack <- A
PUSH 3 //Top of Stack <- immediate value 3
ADD   //Top of Stack <- (A + 3)
PUSH B //Top of Stack <- B
MUL //Top of Stack <- (A + B)*3
PUSH C //Top of Stack <- C
PUSH 5 //Top of Stack <- immediate value 5
MUL //Top of Stack <- C * 5
ADD       //Top of Stack <- (A + B)*3 + c * 5

one address instructions

ADD 3       //AC <- AC + 3
MUL B       //AC <- AC *B
STA T       //M[T] <- AC
LDA C       //AC <- M[C]
MUL 5       //AC <- AC * 5
ADD T       //AC <- AC + M[T]

Here AC is nothing but Register A which is also called as accumulator
T is used to store temporary memory location.

two address instructions

ADD A,#3   //A = A + 3
MUL B       //A = (A + 3) * B
MOV R1,A   //R1= (A + 3) * B
MOV A,#5   //A = 5
MUL C       //A = C*5
ADD A,R1   //A = (A + 3) * B + c * 5

THREE ADDRESS INSTRUCTIONS

ADD A,A,#3   // A= A+3
MUL R1,A,B   // R1= (A+3)*B
MUL C,C,#5   // C= C * 5
ADD A,R1,C   // A= (A+3)*B + c * 5