CODEWITHSUNDEEP
javafunctionif-statementrecursion
Nikhil Gopal
Nikhil Gopal

Reputation: 107

How does this recursive function get to this output?

When I run this piece of code for n=5 the output I get is "5 3 1 1 3 5" I get the 5 3 1 part but after that n=-1 but when I run the code with a debugger it when n=-1 it goes to the line after numbers(n-2);i.e System.out.prt(n+ ""); even though that statement is contained in the if block.

Why does this happen? 

public void numbers(int n)
{
    if(n>0)
    {
        System.out.print(n+" ");
        numbers(n-2);
        System.out.print(n+" ");
    }
}

TLDR : when n=-1 System.out.prt(n+ "");even though it is within the if block which only runs when n>0.

Any help would be much appreciated. Thanks in advance!

Upvotes: 1

Views: 181

Answers (6)

Mohammed Aouf Zouag
Mohammed Aouf Zouag

Reputation: 17132

Here is what happens behind the scenes, for n == 5:

numbers(5);
    if(5 > 0)--> true : 
        System.out.print(5 + " "); // (1)
        numbers(3);
        |   if(3 > 0)--> true : 
        |       System.out.print(3 + " "); // (2)
        |       numbers(1);
        |       |   if(1 > 0)--> true : 
        |       |       System.out.print(1 + " "); // (3)
        |       |       numbers(-1);
        |       |       System.out.print(1 + " "); // (4)
        |       System.out.print(3 + " "); // (5)
        System.out.print(5 + " "); // (6)

Notice how each number is supposed to be printed twice:

System.out.print(n + " "); // print once
numbers(n-2);
System.out.print(n + " "); // print twice

Upvotes: 1

N Jacobs
N Jacobs

Reputation: 341

Remove the final System.out.print(n+" "); After the recursive numbers call it comes back with the original value for n and prints this again.

It bubbles back from the deepest level where it prints 1 two times up to the call with number 3, which is printed again, and also 5 is printed again.

If you want it to update the value after it is printed the first time you will have to update the variable n by performing n-=2 instead of n-2

Upvotes: 1

mohitm
mohitm

Reputation: 163

Follow the recursion stack.

stack representation with and output flow:

             ------------------------->
             5           3            1
num(5)-->  num(3) --> num(1) --> num(-1) --> void
             5           3            1
             <-------------------------

Now since numbers(-1) does not satisfy the if condition, the program control comes out of the if block, and returns void. Start popping out the stack now (recursion): 5 3 1 1 3 5. That's what the output you get, and is expected.

Upvotes: 0

Kousi
Kousi

Reputation: 460

Java maintain method call into stack and as first 5,3,1 gets printed after which method call resume execution for remaining calls and prints 1,3,5 from stack(last call in stack get's picked up first).

Upvotes: 0

Geis
Geis

Reputation: 142

after n=-1, the recursive call will end and return back to the caller method to continue. first commands it found is print.

If you try to debug with the debugger, you will see how rational it is.

Upvotes: 0

David Haim
David Haim

Reputation: 26496

    5 first system.out then number(5-2)
    |
    ----> 3  first system.out then number(5-2)
          |
           ----->1 first system.out then number(5-2)
                 (smaller than 0) , returning
                 |
                 1 second system.out
                 |
           3<----  second system.out
           |
5<---------  second system.out

Upvotes: 0

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