Is there any effective way to round a division?

[rgork]

Round to next integer that is. If the remainders are 0,5<, then it should be rounded down to next integer and when it's 0,5>=, then it should be rounded up to next integer. Say, I have 8593/3384, which is about 2,53. It should be rounded to 3. Then if I have 9/4, which is 2,25, it should be rounded to 2.

I tried to do it somehow like this:

int example(){
int a = 9;
int b = 4;
inc c = 9/4;
int d = 9%4;
double e = d/b;
if (e < 0,5){
return c++;
}
else{
return c;
}
}

But that's such a clumsy way and in many cases, doubles of floats are very inaccurate. Is there any effective way to do this?

[Catfish3]

Maybe these could be useful in your case?

http://cplusplus.com/reference/cmath/modf/
http://cplusplus.com/reference/cmath/ceil/
http://cplusplus.com/reference/cmath/floor/

[keskiverto]

Conversion to integer drops the remainder. Floating point numbers are inaccurate. Therefore, the following is inaccurate too:

int a = 9;
int b = 4;
int c = static_cast<double>( a ) / b + 0.5;

Question is, how often does it fail in practice?

[rgork]

It doesn't need to be 100% accurate, but there doesn't seem to be a way to have it 100%. I guess that will do, thanks.

[keskiverto]

Thinking a bit further, since we have integers:

int c = a / b;
int d = a % b;
int e = b - d;
c += ( d < e ) ? 0 : 1;

That should be accurate for all positive inputs, and does not have if-else branching. Efficiency ... you could test

integer division
integer modulo
integer substraction
conditional expression with integer comparison

vs.

conversion to double
double division
double addition
conversion to int