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decimal rounding

Posted on 1998-12-12
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Last Modified: 2012-05-04
What's the easiest way to round (or truncate) a double value?

example:

I have   3.21453235
I want   3.21

Thanks
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Question by:talaskam
4 Comments
 
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Accepted Solution

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MatthewL earned 50 total points
ID: 1180028
Hi talaskam;

You can shift the number the required number of decimal places, truncate it, and shift it back again.  The function round listed below does that.  It has a limitation on the size of the numbers used ( since I've cast into a long int ).

Matt

#include <iostream.h>
#include <math.h>

double round( double number, int decPlace );


void main( void )
{
      double value = 3.21453235;

      value = round( value, 2 );

      cout << value << endl;

}

double round( double number, int decPlace )
{
      long int temp;

      number *=  pow( 10, decPlace );
      temp = (long int)( number + .5 );
      number = ( double ) temp;

      number /= pow( 10, decPlace );

      return( number );
}
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LVL 84

Expert Comment

by:ozo
ID: 1180029
Now try it with
value = round( -3.21453235, 2);
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LVL 22

Expert Comment

by:nietod
ID: 1180030
To avoid the limitations of 32 bit integers and (I believe) to have the correct (well, most accepted) rounding for negatives, use

number *=  pow( 10, decPlace );
number = floor( number + .5 );
number /= pow( 10, decPlace );
0
 

Expert Comment

by:mmachie
ID: 1180031
iff this is for printing the value you can just use the width and precision stuff for printf.
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