Rounding up in C++

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Hi, was wondering if anyone can inform me how to round up in C++.
Basically making a paint estimator and for example

1 ltr of paint covers 8 square meters

float ceiling = 18.7;

ceiling = 18.7/8;

so now ceiling will be 2.3375

but you cant buy 2.3375 litres worth of paint you need 3, so I wana know how I could program it so that I can round 2.3375 or whatever the data input might be up to the next whole number.

Posted 11-Nov-10 4:23am

WurmInfinity

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5 solutions

Solution 1

Add 0.9999 to ceiling and round that down.

Posted 11-Nov-10 4:32am

Nemanja Trifunovic

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WurmInfinity 11-Nov-10 10:33am

Ceiling will be a user input, I just exampled it as a static float. Also have no idea how to round down help? :D

Solution 3

solved it by making it an int and adding +0.5f

Posted 11-Nov-10 8:56am

WurmInfinity

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Alain Rist 12-Nov-10 2:15am

Works only for positive values: 4.99 -> 5.49 -> 5 OK, -4.99 -> -4.49 -> -4 BAD

Solution 4

Hi,

A correct answer must deal with negative numbers and with the precision of comparisons between double values.

#include <limits>
#include <cmath>
double Round(double val)
{
    const double eps = std::numeric_limits<double>::epsilon();
    return (val <= eps) && (val >= -eps) ? 0 : val > eps ? 
        val - floor(val) - 0.5 >= eps ? ceil(val) : floor(val) :
        val - ceil(val) + 0.5 <= -eps ? floor(val) : ceil(val);
}

A more flexible approach uses the built-in facilities of the iostream Standard Library:

#include <sstream>
#include <iomanip>
double Round(double val, size_t decimal)
{
    std::ostringstream oss;
    oss << std::fixed << std::setprecision(decimal) << val << std::ends;
    std::istringstream(oss.str()) >> val;
    return val;
}

cheers,
AR

Edit: Handle near zero input

Posted 11-Nov-10 21:56pm

Alain Rist

Updated 12-Nov-10 4:11am

v3

Solution 5

Why not just use the CRT ceil() function? That is exactly what it does.

From docs:

double ceil( double x );

The ceil function returns a double value representing the smallest integer that is greater than or equal to x.

Posted 12-Nov-10 4:14am

basementman

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Alain Rist 12-Nov-10 10:21am

ceil(-4.99) == -4. :(

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Aescleal · Accepted Answer · 2010-11-11T06:06:00

It looks like you want to know how many lots of a certain size are in a particular number. I'd be tempted to try something like:

double round_up_to_nearest_lot( double quantity, double lot_size )
{
    unsigned lots = unsigned( quantity / lot_size );
    if( (lots * lot_size) < quantity ) lots++;
    return lot_size * lots;
}

The key here to work out a low order estimate of how many lots minimum you need to cover a particular quantity then check that it's enough. If not bump the number by one and use that many lots.

Just re-reading your problem a bit.. you actually want to find the number of lots (i.e. cans of paint), so you could do something like:

unsigned number_of_lots_in_quantity( double quantity, double lot_size )
{
    unsigned lots = unsigned( quantity / lot_size );
    if( (lots * lot_size) < quantity ) lots++;
    return lots;
}

and you'd call it along the lines of:

unsigned cans_of_paint_required( number_of_lots_in_quantity( area_to_cover, size_of_can ) );

Anyway, hope that's something approaching what you want and they way I've gone about it is understandable. Note that both those functions are going to be rubbish if the number of lots is greater than the size of an integer on your system.

Cheers,

Ash

Rounding up in C++

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