# floor or ceil for integer division

Discussion in 'C++' started by PengYu.UT, Oct 23, 2006.

1. ### PengYu.UTGuest

Hi,

The usually integer division will round the result to the biggest
integet smaller than the float version division.For example, 10/3 = 3.
I'm wondering if there is any easy way to round it to 4 for this case?

Thanks,
Peng

PengYu.UT, Oct 23, 2006

2. ### Steve PopeGuest

int a,b,c;
c = (int) ((double)(a/b) + 0.4999999);

Steve

Steve Pope, Oct 23, 2006

3. ### Ye DafengGuest

I donot understand what you want to say, 10/3=3.33333, so the integer
you need is 3, is that wrong?

Ye Dafeng, Oct 23, 2006
4. ### Ye DafengGuest

hoops, i misunderstand your meaning, sorry!

Ye Dafeng, Oct 23, 2006
5. ### PengYu.UTGuest

Will this introduce errors if a/b=0.50000000000001?
How can I guarantee the conversion between float and int won't
introduce any errors?

Thanks,
Peng

PengYu.UT, Oct 23, 2006
6. ### Ye DafengGuest

if((a%b) != 0)
c = (int)(a/b) + 1;

Ye Dafeng, Oct 23, 2006
7. ### Steve PopeGuest

Well, you only asked that 10/3 round up to 4.
These give you either the nearest value, or they round down.
If you need more exact behavior use something like rint(),
and refer to the man pages for it.

Steve

Steve Pope, Oct 23, 2006
8. ### Jim LangstonGuest

If you don't want to go with the converstion to float, then look at the
remainder.

int Operator = 10;
int Devisor = 3;
int Result = Operator / Devisor;

if ( Operator % Devisor != 0 )
Result++;

This will round up.

if ( Operator % Devisor >= Devisor / 2 )
Result++;
This will round to the closest in cases where the devisor is devisible by 2.

Neither of these handle negatives correctly.

Jim Langston, Oct 23, 2006
9. ### Salt_PeterGuest

you are attempting to introduce a significant error by suggesting that
10/3 should be 4? Thats a 20% error! What happens when a 20% error gets
propagated?
What are the requirements for that 3 to become a 4? and why?
have you looked at modulus? 10%3, which will result 1 as an adjustment?
Although i still can't understand why.

Do you realize that you'ld have to multiply that doubles's error as
displayed by 1.0 x 10^14 in order for it to impact a cast to an integer
value? In fact, since an integer has both a max and min limitation with
a relatively puny and microscopic valid range, that floating number has
an amazing, mind-boggling count of significant figures.
Here, observe what happens to an integer when it reaches its max
allowed value:

#include <iostream>
#include <limits>

int main()
{
int min = std::numeric_limits<int>::min();
int max = std::numeric_limits<int>::max();
std::cout << "range of an integer on this platform:";
std::cout << "\nmin: " << min;
std::cout << "\nmax: " << max;
std::cout << "\nrollover = " << ++max; // rollover

std::cout << std::endl;

return 0;
}

/* your mileage will vary... i'm running a 64 bit platform
range of an integer on this platform:
min: -2147483648
max: 2147483647
rollover = -2147483648 !!!
*/

/*
range of a double on this platform:
min: 2.22507e-308
max: 1.79769e+308
*/
thats around 4e616 !!! care to compare a roundoff error of 1.0e-14 to
that range?

floating numbers are not whole numbers and they do typically roundof
the last digit by 0.5 or so. But lets take that into context here.

Salt_Peter, Oct 23, 2006
10. ### Andrey TarasevichGuest

That's true for positive integers. If one of them is negative though,
the direction of rounding will normally be the opposite (i.e. integer
division normally rounds towards zero).
Increase the dividend by 'divisor - 1' before performing the division:

(10 + 3 - 1) / 3 = 4

Andrey Tarasevich, Oct 23, 2006