CODEWITHSUNDEEP
javaexceptionparameter-passing
skm
skm

Reputation: 25

My codes give an exception while using recursive function

i am trying to print numbers from 1 to 10 without using loops in java. When n+1 is passed to recursivefun method call in line 6,it works fine. But when n++ is passed,the code throws an error :/

public class PrintWithoutUsingLoops {

    public static void recursivefun(int n) {
        if (n <= 10) {
            System.out.println(n);
            recursivefun(n++);//an exception is thrown at this line.
        }
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        recursivefun(1);
    }

}

Upvotes: 1

Views: 246

Answers (4)

Ganji
Ganji

Reputation: 1

Indeed both post and pre increment operators increments the value of a variable. Their behavior changes based on the context of the usage. Assume the following code:

For loop 1: for (i=0; i<10; i++) ...

For loop 2: for (i=0; i<10; ++i) ...

In both of the above statements, the for loop iterates 10 times irrespective of the increment style used. However consider the following code:

int x = 10;
int y = 20;
int z = x++ + ++y; // z = 10 + 21
System.out.println("x = " + x); // prints 11
System.out.println("y = " + y); // prints 21
System.out.println("z = " + z); // prints 31

Hence from your code it is evident that recursivefun(n++);

calls recursivefun with argument 1 infinitely. To avoid StackOverFlow error use either ++n or n+1.

Upvotes: 0

azro
azro

Reputation: 54148

recursivefun(n++);

This line means : call recursivefun(n); and then increment n by 1 so you'll always call your fuction with n=1 and caused a stackOverflow sure

So you need to increment n BEFORE all the function, you have some options : 

recursivefun(n+1);
//-----------------------------
n++;
recursivefun(n);
//-----------------------------
recursivefun(++n); //pre-cincrement

Upvotes: 2

xenteros
xenteros

Reputation: 15842

recursivefun(n++);

is a call with post-increment. Post increment is a mechanism which enlarges the value after it is read. In this case, you always pass 1.

You can use pre-increment which is ++n which first: increments and then passes the value.

The exception you get is StackOverflowError which means, that the stack is full and JVM cannot store more calls on stack, so it won't be able to revert.

Upvotes: 1

Eran
Eran

Reputation: 393851

recursivefun(n++);

passes the original value of n to the recursive call (since you are using the post-increment operator), making this recursion infinite (since each recursive call gets the same value of n, which never reaches 11) and leading to StackOverflowError.

Change it to

recursivefun(n+1);

or

recursivefun(++n);

Upvotes: 6

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