Thread: Formatting Scientific Notation output

Code:

#include <stdio.h>
#include <stdlib.h>
// Example program............. copywright protected, 2003
int main()
{
	float data;

	data = 0.023;

                printf ("e = %e\n",data);

	return 0;
}

for my result I am getting.

e = 2.300000e-002

I would like to get 2.3e-002

but if my data is... data = 0.023900
I want to get 2.39e-002, not 2.390000e-002

How to set the %e formatter so I don't get any trailing zeros.

Thanks.

printf( "e = %.2e\n", data );

will get you 2.39e-002

bump

>>There has to be a way to do that with a different %e format...
Nope, %e prints extra 0's according to the precision of the data type that you're printing. So if the type has 6 digits of precision and you only fill 2 of them, %e gives you 4 0's. You can get around that by finding out what the precision is that you fill and doing this

Code:

printf("%.*e\n", precision, data);

Or you can take more control over your data by converting it to a string and changing what you need to. The biggest problem with the previous solution is finding what the precision should be for the %e format. The next solution doesn't have that problem :-)

Code:

#include <stdio.h>
#include <ctype.h>

void print_e(float value)
{
  char *s;
  char *p;
  char  buff[50] = {0};

  /* Stringize so we can work with it */
  sprintf(buff, "%e", value);

  /* Copy everything except 0's up to the e */
  for (s = p = buff; (*p = *s) != 'e'; s++)
  {
    if (*s != '0')
    {
      p++;
    }
  }

  /* Copy everything after the e */
  while ((*p = *s) != '\0')
  {
    p++;
    s++;
  }

  /* It's printed nicely :-) */
  printf("%s\n", buff);
}

int main()
{
  print_e(0.023f);
  print_e(0.023900f);
  
  return 0;
}