Java Math floor() Method

The floor() method of Math class returns the largest double value that is equal to a mathematical integer and is less than or equal to the argument.

Syntax:

public static double floor(double a)

Parameters:

The parameter ‘a’ represents the value.

Return Value:

The floor () method returns the largest floating-point value that is equal to a mathematical integer and is less than or equal to the argument.

Special cases of the floor() method are as follows:

• It returns the result same as the argument, if the argument passed is equal to an integer or is a NaN or an infinity or positive zero or negative zero value

Example 1:

public class JavaMathFloorExample1 {
    public static void main(String[] args) {
        double a=10.780;
        //returns the largest double value that is less than or equal to the a
        System.out.println("Largest double value of "+ a+" = "+Math.floor(a));
    }
}

Output:

Largest double value 0.0 = 0.0

Example 2:

public class JavaMathFloorExample2 {
    public static void main(String[] args) {
        double a=10.780/0.0d;
        //It returns the result same as 'a', if 'a' is equal to positive infinity
        System.out.println("Largest double value of "+ a+" = "+Math.floor(a));
    }
}

Output:

Largest double value of Infinity = Infinity

Example 3:

public class JavaMathFloorExample3 {
    public static void main(String[] args) {
        double x=-110.780/0.0d;
        //It returns the result same as x, if x is equal to a Negative infinity
        System.out.println("Largest double value of "+ x+" = "+Math.floor(x));
    }
}

Output:

Largest double value of -Infinity = -Infinity

Example 4:

public class JavaMathFloorExample4 {
    public static void main(String[] args) {
        double a=0.0/0.0d;
        //It returns the result same as a, if a is equal to NaN
        System.out.println("Largest double value of "+ a+" = "+Math.floor(a));
    }
}

Output:

Largest double value of NaN = NaN

Example 5:

public class JavaMathFloorExample5 {
    public static void main(String[] args) {
        double a=0.0d;
        //It returns the result same as a, if a is equal to zero
        System.out.println("Largest double value of "+ a+" = "+Math.floor(a));
    }
}

Output:

Largest double value of 0.0 = 0.0