Perl – In Perl, how can I limit the number of places after the decimal point but have no trailing zeroes

Output of perldoc -q round

Does Perl have a round() function? What about ceil() and floor()? Trig functions?

Remember that int() merely truncates toward 0. For rounding to a certain number of digits, sprintf() or printf() is usually the easiest route.

    printf("%.3f", 3.1415926535);       # prints 3.142

The POSIX module (part of the standard Perl distribution) implements ceil(), floor(), and a number of other mathematical and trigonometric functions.

    use POSIX;
    $ceil   = ceil(3.5);                        # 4
    $floor  = floor(3.5);                       # 3

In 5.000 to 5.003 perls, trigonometry was done in the Math::Complex module. With 5.004, the Math::Trig module (part of the standard Perl distribution) implements the trigonometric functions. Internally it uses the Math::Complex module and some functions can break out from the real axis into the complex plane, for example the inverse sine of 2.

Rounding in financial applications can have serious implications, and the rounding method used should be specified precisely. In these cases, it probably pays not to trust whichever system rounding is being used by Perl, but to instead implement the rounding function you need yourself.

To see why, notice how you'll still have an issue on half-way-point alternation:

    for ($i = 0; $i < 1.01; $i += 0.05) { printf "%.1f ",$i}

    0.0 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.4 0.4 0.5 0.5 0.6 0.7 0.7
    0.8 0.8 0.9 0.9 1.0 1.0

Don't blame Perl. It's the same as in C. IEEE says we have to do this. Perl numbers whose absolute values are integers under 2**31 (on 32 bit machines) will work pretty much like mathematical integers. Other numbers are not guaranteed.

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