Velocity Reviews > C++ > floor or ceil for integer division

# floor or ceil for integer division

PengYu.UT@gmail.com
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Posts: n/a

 10-23-2006
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

Steve Pope
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 10-23-2006
http://www.velocityreviews.com/forums/(E-Mail Removed) <(E-Mail Removed)> wrote:

>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?

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

Steve

Ye Dafeng
Guest
Posts: n/a

 10-23-2006
(E-Mail Removed) wrote:
> 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
>

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

Ye Dafeng
Guest
Posts: n/a

 10-23-2006
Ye Dafeng wrote:
> (E-Mail Removed) wrote:
>> 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
>>

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

hoops, i misunderstand your meaning, sorry!

PengYu.UT@gmail.com
Guest
Posts: n/a

 10-23-2006

On Oct 22, 9:53 pm, (E-Mail Removed) (Steve Pope) wrote:
> (E-Mail Removed) <(E-Mail Removed)> wrote:
> >The usually integer division willroundthe 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 toroundit to 4 for this case?int a,b,c;

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

Will this introduce errors if a/b=0.50000000000001?
>
> See also rint(), nearbyint() in <cmath>.

How can I guarantee the conversion between float and int won't
introduce any errors?

Thanks,
Peng

Ye Dafeng
Guest
Posts: n/a

 10-23-2006
(E-Mail Removed) wrote:
>
> On Oct 22, 9:53 pm, (E-Mail Removed) (Steve Pope) wrote:
>> (E-Mail Removed) <(E-Mail Removed)> wrote:
>>> The usually integer division willroundthe 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 toroundit to 4 for this case?int a,b,c;

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

> Will this introduce errors if a/b=0.50000000000001?
>> See also rint(), nearbyint() in <cmath>.

> How can I guarantee the conversion between float and int won't
> introduce any errors?
>
> Thanks,
> Peng
>

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

Steve Pope
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Posts: n/a

 10-23-2006
(E-Mail Removed) <(E-Mail Removed)> wrote:

>On Oct 22, 9:53 pm, (E-Mail Removed) (Steve Pope) wrote:

>> (E-Mail Removed) <(E-Mail Removed)> wrote:

>> >The usually integer division willroundthe 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 toroundit to 4 for this

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

>Will this introduce errors if a/b=0.50000000000001?

Well, you only asked that 10/3 round up to 4.

>> See also rint(), nearbyint() in <cmath>.

>How can I guarantee the conversion between float and int won't
>introduce any errors?

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

Jim Langston
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Posts: n/a

 10-23-2006
<(E-Mail Removed)> wrote in message
news:(E-Mail Removed) ups.com...
> 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

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.

Salt_Peter
Guest
Posts: n/a

 10-23-2006
(E-Mail Removed) wrote:
> On Oct 22, 9:53 pm, (E-Mail Removed) (Steve Pope) wrote:
> > (E-Mail Removed) <(E-Mail Removed)> wrote:
> > >The usually integer division willroundthe 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 toroundit to 4 for this case?int a,b,c;

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

> Will this introduce errors if a/b=0.50000000000001?
> >
> > See also rint(), nearbyint() in <cmath>.

> How can I guarantee the conversion between float and int won't
> introduce any errors?
>
> Thanks,
> Peng

Your worried about an insignificant, tiny, non-measureable error whilst
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 !!!
*/

Now observe the unadulterated double:

/*
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.

Andrey Tarasevich
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Posts: n/a

 10-23-2006
(E-Mail Removed) wrote:
> ...
> The usually integer division will round the result to the biggest
> integet smaller than the float version division.For example, 10/3 = 3.

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).

> I'm wondering if there is any easy way to round it to 4 for this case?

Increase the dividend by 'divisor - 1' before performing the division:

(10 + 3 - 1) / 3 = 4

--
Best regards,
Andrey Tarasevich