method

round

v1_9_3_392 - Show latest stable - 0 notes - Class: Float

round(p1 = v1) public

Rounds flt to a given precision in decimal digits (default 0 digits). Precision may be negative. Returns a floating point number when ndigits is more than zero.

1.4.round      #=> 1
1.5.round      #=> 2
1.6.round      #=> 2
(-1.5).round   #=> -2

1.234567.round(2)  #=> 1.23
1.234567.round(3)  #=> 1.235
1.234567.round(4)  #=> 1.2346
1.234567.round(5)  #=> 1.23457

34567.89.round(-5) #=> 0
34567.89.round(-4) #=> 30000
34567.89.round(-3) #=> 35000
34567.89.round(-2) #=> 34600
34567.89.round(-1) #=> 34570
34567.89.round(0)  #=> 34568
34567.89.round(1)  #=> 34567.9
34567.89.round(2)  #=> 34567.89
34567.89.round(3)  #=> 34567.89

Show source

static VALUE
flo_round(int argc, VALUE *argv, VALUE num)
{
    VALUE nd;
    double number, f;
    int ndigits = 0;
    int binexp;
    enum {float_dig = DBL_DIG+2};

    if (argc > 0 && rb_scan_args(argc, argv, "01", &nd) == 1) {
        ndigits = NUM2INT(nd);
    }
    if (ndigits < 0) {
        return int_round_0(flo_truncate(num), ndigits);
    }
    number  = RFLOAT_VALUE(num);
    if (ndigits == 0) {
        return dbl2ival(number);
    }
    frexp(number, &binexp);

/* Let `exp` be such that `number` is written as:"0.#{digits}e#{exp}",
   i.e. such that  10 ** (exp - 1) <= |number| < 10 ** exp
   Recall that up to float_dig digits can be needed to represent a double,
   so if ndigits + exp >= float_dig, the intermediate value (number * 10 ** ndigits)
   will be an integer and thus the result is the original number.
   If ndigits + exp <= 0, the result is 0 or "1e#{exp}", so
   if ndigits + exp < 0, the result is 0.
   We have:
        2 ** (binexp-1) <= |number| < 2 ** binexp
        10 ** ((binexp-1)/log_2(10)) <= |number| < 10 ** (binexp/log_2(10))
        If binexp >= 0, and since log_2(10) = 3.322259:
           10 ** (binexp/4 - 1) < |number| < 10 ** (binexp/3)
           floor(binexp/4) <= exp <= ceil(binexp/3)
        If binexp <= 0, swap the /4 and the /3
        So if ndigits + floor(binexp/(4 or 3)) >= float_dig, the result is number
        If ndigits + ceil(binexp/(3 or 4)) < 0 the result is 0
*/
    if (isinf(number) || isnan(number) ||
        (ndigits >= float_dig - (binexp > 0 ? binexp / 4 : binexp / 3 - 1))) {
        return num;
    }
    if (ndigits < - (binexp > 0 ? binexp / 3 + 1 : binexp / 4)) {
        return DBL2NUM(0);
    }
    f = pow(10, ndigits);
    return DBL2NUM(round(number * f) / f);
}

round

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