C++ double value rounding problem.....

[alexa12345]

For Input
4
15.01
15
3
3.01

The value of 'avg' is becoming 9.00. I don't understand why this should be 9.01. Please help.

#include<stdio.h>
#include<math.h>

double round(double x) 
{ 	
	return floor(x * pow(10.0,2.0) + 0.5)/pow(10.0,2.0); 
}

int main()
{
	//freopen("10137.txt","r",stdin);
	//freopen("out1.txt","w",stdout);

	long i,n;
	double sum,avg,sum1,a[1005];

	while(scanf("%ld",&n) == 1 && n)
	{
		sum = 0.0;
		for(i = 0 ; i < n ; i++)
		{
			scanf("%lf",&a[i]);
			sum += a[i];
		}
		
		avg = round(sum/n);

		sum1 = sum = 0.0;
		for(i = 0 ; i < n ; i++)
		{
			if(a[i] > avg)
				sum += a[i] - avg;
			else
				sum1 += avg - a[i];
		}

		if(sum < sum1)
			printf("$%0.2lf\n",sum);
		else
			printf("$%0.2lf\n",sum1);
	
	}
	return 0;
}

[poncho4all]

have you tried n=0.0 or making it a double too

[NathanOliver]

well when i add you number together and i dived by 5 I'm getting 8.004. so with your round function you would have ((8.004 * 100) +.5) / 100 = (8.4 + .5) / 100 = 8.9 / 100 = 8.009. with the floor of 8.009 being 8. I'm not sure why you are doing this though. also with double values the last digit can become a one because it is an approximation of the number unlike an int where it is the number.

[alexa12345]

my intention was to compare two double numbers. Is there any better ideas how I can do that? Because you cant do just
if(a>b) then.....
where a and b are both double.

[mvmalderen]

my intention was to compare two double numbers. Is there any better ideas how I can do that? Because you cant do just
if(a>b) then.....
where a and b are both double.

I guess you'll find this useful:
http://www.parashift.com/c++-faq-lite/newbie.html#faq-29.17