Question

Write a program to evaluate the arithmetic statement :
x=a-b+c*(d*e-f)/g+h*k
a) using a general register computer with two address instructions.

Answer 1

Answer:

X=A-B+C*(D*E-F)/G+H*k a. ONE ADDRESS INSTRUCTIONS: LOAD A AC<-M[A] SUB B AC<-AC-M[B] STORE T M[T]<-AC LOAD D AC<-M[D] MUL E AC<-AC*M[E] SUB F AC<-AC-M[F] MUL C AC<-AC*M[C] ADD T AC<-AC+M[T] STORE T M[T]<-AC LOAD H AC<-M[H] MUL K AC<-AC*M[K] ADD G AC<-AC+M[G] STORE T1 M[T1]<-AC LOAD T AC<-M[T] DIV T1 AC<-AC/M[T1] STORE X M[X]<-AC b.ZERO ADDRESS INSTRUCTIONS: RPN:AB-CDE*F-*GHK*+/ PUSH A TOS<-A PUSH B TOS<-B SUB...

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Answer 2

Answer:

Here is a program in assembly language that evaluates the arithmetic statement "x = a - b + c * (d * e - f) / g + h * k" using a general register computer with two address instructions:

   load a, R1

   load b, R2

   sub R1, R2, R3

   load c, R4

   load d, R5

   load e, R6

   mul R5, R6, R7

   load f, R8

   sub R7, R8, R9

   load g, R10

   div R9, R10, R11

   mul R4, R11, R12

   add R3, R12, R13

   load h, R14

   load k, R15

   mul R14, R15, R16

   add R13, R16, R17

   store R17, x

Explanation:

This program uses the following two-address instructions:

load: loads a value from memory into a register

store: stores a value from a register into memory

add: adds the contents of two registers and stores the result in a third register

sub: subtracts the contents of two registers and stores the result in a third register

mul: multiplies the contents of two registers and stores the result in a third register

div: divides the contents of two registers and stores the result in a third register

Note: This program assumes that the values of a, b, c, d, e, f, g, h and k are already stored in memory and the memory address for x is available to store the result. Also, the program uses R1 to R17 as registers.

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