Why does this happen in Java? With two equal numbers

To get the output `10` in both cases, we need to understand how floating-point precision and rounding work in Java. When dealing with floating-point numbers, the precision is limited, which can lead to unexpected results. The `double` type in Java follows the IEEE 754 standard for floating-point arithmetic, which means it can only accurately represent a certain number of decimal places. The numbers you provided have 16 and 15 decimal places respectively, which causes issues due to precision limits. To ensure both cases output `10`, you can use the `BigDecimal` class in Java. `BigDecimal` provides arbitrary-precision arithmetic and can handle very large and very small floating-point numbers with great accuracy. You can use `BigDecimal` to perform the floor operation. Here's the modified code to achieve this:

import java.math.BigDecimal;
import java.math.RoundingMode;

public class Main {
    public static void main(String[] args) {
        // Reason 16 decimals  ->> Output: 10
        BigDecimal num1 = new BigDecimal("10.9999999999999999");
        Integer res1 = num1.setScale(0, RoundingMode.FLOOR).intValue();
        System.out.println(res1);

        // Reason 15 decimals  ->> Output: 10
        BigDecimal num2 = new BigDecimal("10.999999999999999");
        Integer res2 = num2.setScale(0, RoundingMode.FLOOR).intValue();
        System.out.println(res2);
    }
}

In this code: 1. `BigDecimal` is used to create a floating-point number with high precision. 2. `setScale(0, RoundingMode.FLOOR)` is used to scale the number to 0 decimal places, effectively performing the floor operation. 3.intValue() converts the resulting `BigDecimal` to an integer.