Round a Decilam withOUT the leading zero

[ericthered13]

Does anyone know how I could use printf to display a decimal number without the leading zero?

Something like,
{
...
x = .45;
printf ("%lf", x);
...
}
and have that display '.45' instead of '0.45'?

[redhead]

You can't not without some black magic, I think the printf() man page explains it quite well

f, F
The floating arg is converted to decimal notation in the style [-]ddd.ddd, where the number of fractional digits is equal to the precision specification. If the precision is missing, it is taken as 6; if the precision is zero and the # flag is not specified, no decimal point character appears. If a decimal point character appears, at least one digit appears before it. The value is rounded to the appropriate number of digits.

[redhead]

By black magic I'm thinking of something like this:

/*
 * bprint() converts a given double value in a way so
 * any value less than 1 will _not_ have leading zeroes
 *
 * buffer     - an allocated char* 
 *              where the return value will be stored
 * value      - the value intended for conversion
 * precission - the wanted precission setting, as
 *              described for any *printf function
 */

#define LENGTH 256

int bprint(char buffer[], double value, unsigned long int precission)
{
  char format[LENGTH +1], res[LENGTH +1];
  int i=0, j=0;
#ifdef HAVE_SNPRINTF
  /* create a format required by the user */
  if(!snprintf(format, LENGTH, "%%.%dlf", prescission))
    return -1;
  /* fill in the value in our prefered format */
  if(!snprintf(res, LENGTH, format, value))
    return 0;
#else
  if(!sprintf(format, "%%.%dlf", precission))
    return -1;
  if(!sprintf(res, format, value))
    return 0;
#endif
  if(value < 1.0)
    while(res[i]&&res[i]=='0')
      ++i; /* skip any leading zeroes */
  for(;res[i];++i,++j)
    buffer[j]=res[i]; /* copy rest of value */
  /* nice NULL termination */
  buffer[j]='\0';
  return 1;
}