C# Math.ILogB() – Syntax & Examples

C# Math.ILogB() – Examples

In this tutorial, we will learn about the C# Math.ILogB() method, and learn how to use this method to compute base 2 integer logarithm of a specified number, with the help of examples.

ILogB(Double)

Math.ILogB(x) returns the base 2 integer logarithm of a specified number x.

Syntax

The syntax of ILogB() method is

Math.ILogB(Double x)

where

Parameter	Description
x	The number whose logarithm is to be found.

Return Value

The method returns Int32 value.

Example 1 – ILogB(x)

In this example, we will take some double-precision floating-point numbers and find their integer logarithm value with base value of 2, using Math.ILogB() method.

C# Program

using System;
 
class Example {
    static void Main(string[] args) {
        Double x;
        Double result;

        x = 2;
        result = Math. ILogB(x);
        Console.WriteLine($"ILogB({x}) = {result}");

        x = 3;
        result = Math. ILogB(x);
        Console.WriteLine($"ILogB({x}) = {result}");

        x = 4;
        result = Math. ILogB(x);
        Console.WriteLine($"ILogB({x}) = {result}");
    }
}

Output

ILogB(2) = 1
ILogB(3) = 1
ILogB(4) = 2

Please note that ILogB() returns an integer value. Therefore, the precision or decimals points are trimmed out after computing the logarithm.

Conclusion

In this C# Tutorial, we have learnt the syntax of C# Math.ILogB() method, and also learnt how to use this method with the help of C# example programs.