Question

Your code must approximate the square root of an integer between 0 and 2^31-1 Using integer maths is sufficient (no floating point or fractions are required); your code will return the truncated (integer portion) of the square root. Your code must be in an assembly language subroutine which is called by a C function for testing. Be sure to use registers according to the ARM calling convention.

Base your software on the following pseudocode:

Approximate square root with bisection method
INPUT: Argument x, endpoint values a, b, such that a < b OUTPUT: value which differs from sqrt(x) by less than 1

done = 0
a=0
b = square root of largest possible argument (e.g. ~216). c = -1
do {

c_old <- c
c <- (a+b)/2
if (c*c == x) {

               done = 1
        } else if (c*c < x) {

a <- c } else {

b <- c

}
} while (!done) && (c != c_old)

return c

Write an assembly code subroutine to approximate the square root of an
argument using the bisection method. All math is done with integers, so the
resulting square root will also be an integer

GOOD LUCK!
*----------------------------------------------------------------------------*/

__asm int my_sqrt(int x){
  
   //Write your code here

}

/*----------------------------------------------------------------------------
MAIN function
*----------------------------------------------------------------------------*/
int main(void){
   volatile int r, j=0;
   int i;
r = my_sqrt(0); // should be 0
r = my_sqrt(25); // should be 5
   r = my_sqrt(133);    // should be 11
for (i=0; i<10000; i++){
       r = my_sqrt(i);
j+=r;
}
   while(1)
       ;
}

Answer 1

#include <stdio.h>

#include <math.h>

int my_sqrt(int x){

int ivar; //taking integer

float fvar; //taking float

fvar=sqrt((double)x); //calculating root of number

ivar=fvar; //converting float to integer

if(ivar==fvar) // if float and int both are same then it is perfect square

return ivar; //returning root

else

return 0;

}

int main(){

int r, j=0;

int i;

r = my_sqrt(0); // should be 0

r = my_sqrt(25); // should be 5

r = my_sqrt(133); // should be 11

for (i=0; i<10000; i++){

r = my_sqrt(i);

j+=r;

}

printf("%d",j);

return 0;

}