Avoid trailing zeroes in printf()

c, printf

Solution

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

Problem

I keep stumbling on the format specifiers for the printf() family of functions. What I want is to be able to print a double (or float) with a maximum given number of digits after the decimal point. If I use: ``` printf("%1.3f", 359.01335); printf("%1.3f", 359.00999); ``` I get ``` 359.013 359.010 ``` Instead of the desired ``` 359.013 359.01 ``` Can anybody help me?

Original source