MathHelper.floorDiv

Returns the largest (closest to positive infinity) {@code int} value that is less than or equal to the algebraic quotient. There is one special case, if the dividend is the {@linkplain Integer#MIN_VALUE Integer.MIN_VALUE} and the divisor is {@code -1}, then integer overflow occurs and the result is equal to {@code Integer.MIN_VALUE}. <p> Normal integer division operates under the round to zero rounding mode (truncation). This operation instead acts under the round toward negative infinity (floor) rounding mode. The floor rounding mode gives different results from truncation when the exact result is negative. <ul> <li>If the signs of the arguments are the same, the results of {@code floorDiv} and the {@code /} operator are the same. <br> For example, {@code floorDiv(4, 3) == 1} and {@code (4 / 3) == 1}.</li> <li>If the signs of the arguments are different, the quotient is negative and {@code floorDiv} returns the integer less than or equal to the quotient and the {@code /} operator returns the integer closest to zero.<br> For example, {@code floorDiv(-4, 3) == -2}, whereas {@code (-4 / 3) == -1}. </li> </ul>

@param x the dividend @param y the divisor @return the largest (closest to positive infinity) {@code int} value that is less than or equal to the algebraic quotient. @throws ArithmeticException if the divisor {@code y} is zero @see #floorMod(int, int) @see #floor(double)

  1. int floorDiv(int x, int y)
    class MathHelper
    static
    int
    floorDiv
    (
    int x
    ,
    int y
    )
  2. long floorDiv(long x, int y)
  3. long floorDiv(long x, long y)

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