List of utility methods to do floor
double | floor(double a) Returns the largest (closest to positive infinity) double value that is less than or equal to the argument and is equal to a mathematical integer. return Math.floor(a);
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double | floor(double a, double precision) Returns a rounded number smaller than a with a defined precision. if (precision == 0.0) { return 0.0; return Math.floor(a / precision) * precision; |
double | floor(double a, int cutOfDigits) Modification of Math#floor(double) taking an additional argument representing the requested accuracy in the following way: double fac = Math.pow(10, digits);
double fac = Math.pow(10, cutOfDigits); return fac * Math.floor(a / fac); |
double | floor(double a, int n) floor double p = Math.pow(10.0, n); return Math.floor((a * p) + 0.5) / p; |
int | floor(double d) floor return (int) (d + FLOOR) - INT; |
int | floor(double d) floor return (int) Math.floor(d); |
int | floor(double d) Unchecked implementation to round a number down. int i = (int) d; return d < i ? i - 1 : i; |
double | floor(double d, int exp) This method returns the largest double value that is smaller than d = x * 10exp where x is rounded down to the closest integer.
double x = 1.0 * Math.pow(10.0, (double) exp); return Math.floor(d / x) * x; |
double | floor(double d, int p) floor long tmp = (long) Math.pow(10, p); double num = Math.floor(d * tmp); num /= tmp; return num; |
int | floor(double num) floor final int numInt = (int) num; return numInt == num ? numInt : numInt - (int) (Double.doubleToRawLongBits(num) >>> 63); |