Raise x to the power n using Recursion in Java

The power of a number can be calculated using recursion. Here the number is x and it is raised to power n.

A program that demonstrates this is given as follows:

Example

 Live Demo

public class Demo {
   static double pow(double x, int n) {
      if (n != 0)
         return (x * pow(x, n - 1));
      else
         return 1;
   }
   public static void main(String[] args) {
      System.out.println("7 to the power 3 is " + pow(7, 3));
      System.out.println("4 to the power 1 is " + pow(4, 1));
      System.out.println("9 to the power 0 is " + pow(9, 0));
   }
}

Output

7 to the power 3 is 343.0
4 to the power 1 is 4.0
9 to the power 0 is 1.0

Now let us understand the above program.

The method pow() calculates x raised to the power n. If n is not 0, it recursively calls itself and returns x * pow(x, n - 1). If n is 0, it returns 1. A code snippet which demonstrates this is as follows:

static double pow(double x, int n) {
   if (n != 0)
      return (x * pow(x, n - 1));
   else
      return 1;
}

In main(), the method pow() is called with different values. A code snippet which demonstrates this is as follows:

public static void main(String[] args) {
   System.out.println("7 to the power 3 is " + pow(7, 3));
   System.out.println("4 to the power 1 is " + pow(4, 1));
   System.out.println("9 to the power 0 is " + pow(9, 0));
}