shoes - 1 year ago 67

Java Question

I am required to write a recursive method. I have written the code to perform the task without recursion. I am getting the error **Exception in thread "main" java.lang.StackOverflowError at Exercise13r.recursion(Exercise13r.java:29)**. Code is... to enter a number then if result is even, divide by 2, if result is odd, multiply by 3 and subtract 1. Obviously I am looping but not sure why. Any assistance would be appreciated.

`import java.util.Scanner;`

public class Exercise13r

{

public static void main(String[] args)

{

// Initialize variables

long number = 0;

Scanner in = new Scanner(System.in);

System.out.println ("Enter a starting number: ");

number = in.nextInt ();

System.out.println ("Your starting number is: " + number);

if (number != 1)

{

recursion(number);

}

}

public static void recursion(long n)

{

if (n % 2 == 0)

{

recursion(n/2);

}

else

{

recursion(n*3-1);

}

System.out.println ("number: " + n);

return;

}

}

Answer Source

Your base case `if (number != 1)`

needs to be inside the definition of the function so that it actually knows when to stop. Right now your program eventually reduces to calling `recursion(1)`

and your function will still call itself recursively (what else can it do?) so it ends up calling `recursion(2)`

which leads to `recursion(1)`

again and so on.

Note that this becomes apparent if you move `System.out.println ("number: " + n);`

*before* the recursive calls. Since you have infinite recursion it never gets around to printing anything preventing you from seeing the problem.

Here is a minimal working example:

```
class Exercise13r {
public static void main(String[] args) {
recursion(12);
}
public static void recursion(long n) {
System.out.println ("number: " + n);
if (n != 1) {
if (n % 2 == 0) {
recursion(n/2);
} else {
recursion(n*3-1);
}
}
}
}
```

Output:

```
number: 12
number: 6
number: 3
number: 8
number: 4
number: 2
number: 1
```