Avoid trailing zeroes in printf()

This can't be done with the normal printf format specifiers. The closest you could get would be:

printf("%.6g", 359.013); // 359.013
printf("%.6g", 359.01);  // 359.01

but the ".6" is the total numeric width so

printf("%.6g", 3.01357); // 3.01357

breaks it.

What you can do is to sprintf("%.20g") the number to a string buffer then manipulate the string to only have N characters past the decimal point.

Assuming your number is in the variable num, the following function will remove all but the first N decimals, then strip off the trailing zeros (and decimal point if they were all zeros).

char str[50];
sprintf (str,"%.20g",num);  // Make the number.
morphNumericString (str, 3);
:    :
void morphNumericString (char *s, int n) {
    char *p;
    int count;

    p = strchr (s,'.');         // Find decimal point, if any.
    if (p != NULL) {
        count = n;              // Adjust for more or less decimals.
        while (count >= 0) {    // Maximum decimals allowed.
             count--;
             if (*p == '\0')    // If there's less than desired.
                 break;
             p++;               // Next character.
        }

        *p-- = '\0';            // Truncate string.
        while (*p == '0')       // Remove trailing zeros.
            *p-- = '\0';

        if (*p == '.') {        // If all decimals were zeros, remove ".".
            *p = '\0';
        }
    }
}

If you're not happy with the truncation aspect (which would turn 0.12399 into 0.123 rather than rounding it to 0.124), you can actually use the rounding facilities already provided by printf. You just need to analyse the number before-hand to dynamically create the widths, then use those to turn the number into a string:

#include <stdio.h>

void nDecimals (char *s, double d, int n) {
    int sz; double d2;

    // Allow for negative.

    d2 = (d >= 0) ? d : -d;
    sz = (d >= 0) ? 0 : 1;

    // Add one for each whole digit (0.xx special case).

    if (d2 < 1) sz++;
    while (d2 >= 1) { d2 /= 10.0; sz++; }

    // Adjust for decimal point and fractionals.

    sz += 1 + n;

    // Create format string then use it.

    sprintf (s, "%*.*f", sz, n, d);
}

int main (void) {
    char str[50];
    double num[] = { 40, 359.01335, -359.00999,
        359.01, 3.01357, 0.111111111, 1.1223344 };
    for (int i = 0; i < sizeof(num)/sizeof(*num); i++) {
        nDecimals (str, num[i], 3);
        printf ("%30.20f -> %s\n", num[i], str);
    }
    return 0;
}

The whole point of nDecimals() in this case is to correctly work out the field widths, then format the number using a format string based on that. The test harness main() shows this in action:

  40.00000000000000000000 -> 40.000
 359.01335000000000263753 -> 359.013
-359.00999000000001615263 -> -359.010
 359.00999999999999090505 -> 359.010
   3.01357000000000008200 -> 3.014
   0.11111111099999999852 -> 0.111
   1.12233439999999995429 -> 1.122

Once you have the correctly rounded value, you can once again pass that to morphNumericString() to remove trailing zeros by simply changing:

nDecimals (str, num[i], 3);

into:

nDecimals (str, num[i], 3);
morphNumericString (str, 3);

(or calling morphNumericString at the end of nDecimals but, in that case, I'd probably just combine the two into one function), and you end up with:

  40.00000000000000000000 -> 40
 359.01335000000000263753 -> 359.013
-359.00999000000001615263 -> -359.01
 359.00999999999999090505 -> 359.01
   3.01357000000000008200 -> 3.014
   0.11111111099999999852 -> 0.111
   1.12233439999999995429 -> 1.122

To get rid of the trailing zeros, you should use the "%g" format:

float num = 1.33;
printf("%g", num); //output: 1.33

After the question was clarified a bit, that suppressing zeros is not the only thing that was asked, but limiting the output to three decimal places was required as well. I think that can't be done with sprintf format strings alone. As Pax Diablo pointed out, string manipulation would be required.

My idea is to calculate the required precision that would not result in trailing zeroes for a given double value and pass it to the "%1.*f" format in printf(). This can even be done as one-liner:

int main() {
    double r=1234.56789;
    int precision=3;
    printf(L"%1.*f", prec(r, precision), r);
}

int prec(const double& r, int precision)
{
    double rPos = (r < 0)? -r : r;
    double nkd = fmod(rPos, 1.0); // 0..0.99999999
    int i, ex10 = 1;
    for (i = 0; i < precision; ++i)
        ex10 *= 10;
    int nki = (int)(nkd * ex10 + 0.5);

    // "Eliminate" trailing zeroes
    int requiredPrecision = precision;
    for (; requiredPrecision && !(nki % 10); )  {
        --requiredPrecision;
        nki /= 10;
    }
    return requiredPrecision;        
}

And here is another %g solution. You should always provide a format precision that is "wide enough" (default is only 6) and round the value. I think this is a nice way to do it:

double round(const double &value, const double& rounding)  {
    return rounding!=0 ? floor(value/rounding + 0.5)*rounding : value;
}

printf("%.12g" round(val, 0.001)); // prints up to 3 relevant digits

Here is my first try at an answer:

void
xprintfloat(char *format, float f)
{
  char s[50];
  char *p;

  sprintf(s, format, f);
  for(p=s; *p; ++p)
    if('.' == *p) {
      while(*++p);
      while('0'==*--p) *p = '\0';
    }
  printf("%s", s);
}

Known bugs: Possible buffer overflow depending on format. If "." is present for other reason than %f wrong result might happen.

Slight variation on above:

1. Eliminates period for case (10000.0).
2. Breaks after first period is processed.

Code here:

void EliminateTrailingFloatZeros(char *iValue)
{
  char *p = 0;
  for(p=iValue; *p; ++p) {
    if('.' == *p) {
      while(*++p);
      while('0'==*--p) *p = '\0';
      if(*p == '.') *p = '\0';
      break;
    }
  }
}

It still has potential for overflow, so be careful ;P

I would say you should use printf("%.8g",value);

If you use "%.6g" you will not get desired output for some numbers like.32.230210 it should print 32.23021 but it prints 32.2302

Hit the same issue, double precision is 15 decimal, and float precision is 6 decimal, so I wrote to 2 functions for them separately

#include <stdio.h>
#include <math.h>
#include <string>
#include <string.h>

std::string doublecompactstring(double d)
{
    char buf[128] = {0};
    if (isnan(d))
        return "NAN";
    sprintf(buf, "%.15f", d);
    // try to remove the trailing zeros
    size_t ccLen = strlen(buf);
    for(int i=(int)(ccLen -1);i>=0;i--)
    {
        if (buf[i] == '0')
            buf[i] = '\0';
        else
            break;
    }

    return buf;
}

std::string floatcompactstring(float d)
{
    char buf[128] = {0};
    if (isnan(d))
        return "NAN";
    sprintf(buf, "%.6f", d);
    // try to remove the trailing zeros
    size_t ccLen = strlen(buf);
    for(int i=(int)(ccLen -1);i>=0;i--)
    {
        if (buf[i] == '0')
            buf[i] = '\0';
        else
            break;
    }

    return buf;
}

int main(int argc, const char* argv[])
{
    double a = 0.000000000000001;
    float  b = 0.000001f;

    printf("a: %s\n", doublecompactstring(a).c_str());
    printf("b: %s\n", floatcompactstring(b).c_str());
    return 0;
}

output is

a: 0.000000000000001
b: 0.000001

I needed that and the first answer from paxdiablo does the trick. But I was not needing truncating and the version below is maybe slightly faster? Starting to search end of string (EOS) after the ".", only one placement of EOS.

//https://stackoverflow.com/questions/277772/avoid-trailing-zeroes-in-printf
//adapted from paxdiablo (removed truncating)
char StringForDouble[50];
char *PointerInString;
void PrintDouble (double number) {
  sprintf(StringForDouble,"%.10f",number); // convert number to string
  PointerInString=strchr(&StringForDouble[0],'.'); // find decimal point, if any
  if(PointerInString!=NULL) {
    PointerInString=strchr(&PointerInString[0],'\0'); // find end of string
    do{
      PointerInString--;
    } while(PointerInString[0]=='0'); // remove trailing zeros
    if (PointerInString[0]=='.') { // if all decimals were zeros, remove "."
      PointerInString[0]='\0';
    } else {
      PointerInString[1]='\0'; //otherwise put EOS after the first non zero char
    }
  }
  printf("%s",&StringForDouble[0]);
}