Thread: display double w/o trailing zeros?

I'm trying to display a list of doubles with out the trailing zeros.
This snipet is just the declaration and printf...the actual code is fine
I just need to modify these intergers for the printf

Code:

   double sq = 0, bb = 0, ac4 = 0, z = 0;
   long sq2 = 0, x1 = 0, x2 = 0;
   printf("%d+%di, %d-%di.\n\n", x1, sq, x1, sq);

This is part of a quadradic solver. I'm inputting x^2+2x+2
and keep getting:

1.000000+1.0000000i, 1.0000000-1.0000000i

instead of;

1+1i, 1-1i

could anyone assist with a method to truncate the trailing zeros?

Code:

   double sq = 0, bb = 0, ac4 = 0, z = 0;
   double sq2 = 0, x1 = 0, x2 = 0;
   printf("%f+%fi, %f-%fi.\n\n", x1, sq, x1, sq);

However, the %d was also producing the same result.
Probly my complier.

still getting output w/ trailing zeros

any idea how to truncate the trailing zeros from a float
in printf?

If you want no decimal digits to be displayed, print as an int:

Code:

double x = 5.2;
printf("%d\n", (int)x);

Output:

Code:

5

I like %g.

For comparison.

Code:

#include <stdio.h>

int main ( void )
{
   int i;
   for ( i = 1; i < 5; ++i )
   {
      double d = 1.0 / i;
      printf ( "d = %g (%%g) = %f (%%f) = %.0f (%%.0f) = %.2f (%%.2f)\n", 
               d, d, d, d );
   }
   return 0;
}

/* my output
d = 1 (%g) = 1.000000 (%f) = 1 (%.0f) = 1.00 (%.2f)
d = 0.5 (%g) = 0.500000 (%f) = 0 (%.0f) = 0.50 (%.2f)
d = 0.333333 (%g) = 0.333333 (%f) = 0 (%.0f) = 0.33 (%.2f)
d = 0.25 (%g) = 0.250000 (%f) = 0 (%.0f) = 0.25 (%.2f)
*/