Question

Spring 2020 CSCI 2450: HW5 (Programming Assignment (10 points) Answer any ONE of the following problems. If you answer both of them, you will get bonus points. Use any instructions and Irvine procedures we covered till now. These problems might require instructions from chapter 07. Problem 1: Greatest Common Divisor ( GCD) Greatest Common Divisor (GCD): The greatest common divisor (GCD) of two integers is the largest integer that will evenly divide both integers. The GCD algorithm involves integer division in a loop, described by the following pseudocode: int GCD (int x, int y) { x = abs(x) labsolute value y = abs(y) do { int n= x % y } while (y > 0) return Implement this function in assembly language and write a test program that calls the function several times, passing it different values. Display all results on the screen.

Answer 1

int main()
{
int x, y, out;
printf("Please enter two integers: \n");
scanf("%d%d", &x, &y);
out = GCD(x,y);
printf("C: Greatest common divisor of two numbers %d and
%d is %d.\n\n", x, y, out);
out = ASMGCD(x,y);
printf("ASM Greatest common divisor of two numbers %d and %d is %d.\n\n", x, y, out);
return 0;
}

int ASMGCD(int x, int y)
{
int n;
__asm {
IF01:
mov eax, x;
cmp eax, 0;
jg ENDIF01;
neg eax; // absolute value
mov x, eax;
ENDIF01:
IF02:
mov eax, y;
cmp eax, 0;
jg ENDIF02;
neg eax; // absolute value
mov y, eax;
ENDIF02:
DO01:
mov eax, x; // numerator
mov ebx, y; // a reg used for denom
sub edx, edx; // zero dx
idiv ebx; // signed integer divide
mov n, edx; // remainder
mov ebx, y;
xchg ebx, x; // x = y
mov ebx, n;
xchg ebx, y; // y = n;
mov ebx, y;
cmp y, 0;
jg DO01; // while (y > 0)
DOEND01:
}
return x;
}