/* compare.c		*/
/* author: Weiwei Chen	*/
/* date: Oct. 11. 2010	*/

#include <stdio.h>

int main(void)
{
	/* variable definitions	*/
	float x = 0, y = 0;
	float r = 0, R = 0;
	float pi = 3.14159;
	float epsilon = 1e-5;
	float diff;

	/* value assigment		*/
	r = 3.2;
	R = 4.2;
	
	/* mathematically, the value of x should be equal to the value of y*/
	x = pi * (R * R - r * r);
	y = pi * R * R - pi * r * r;

	diff = x - y;

	/* however, since the value in computer is approximate		*/
	if(x == y)	/* this condition is false	*/
	{
		printf("x == y \n");
	}
	if(x != y)	/* this condition will be true	*/
	{
		printf("x != y \n");		
	}

	/* how to get the correct answer with the consideration of approximation	*/
	if(diff > -epsilon && diff < epsilon)
	{
		printf("approximately in computer, x == y \n");
	}
	
	if(!(diff > -epsilon && diff < epsilon))
	{
		printf("approximately in computer, x != y \n");
	}
	return 0;
}