CODEWITHSUNDEEP
java
Kelly Tan
Kelly Tan

Reputation: 93

Recursion - Java Programming

I have been trying to figure out why the output is 321 and not 123. I have tried multiple times and looking out lecture slides but I still can't get the solution to this question. 

public static void main(String[] args) {
     printNumber(1);
}

public static void printNumber(int n) {
    if(n!=4) {
         printNumber(n + 1);
         System.out.print(n);
    }
}

Upvotes: 2

Views: 132

Answers (3)

Oghli
Oghli

Reputation: 2340

The reason for this result that you have placed the recursion call statement printNumber(n + 1) before print statement System.out.print(n) so it will keep recursivly calling it until it reach 4 after that it will print numbers from last call n+1 to n as in our example from 3 to 1

try this :

public static void printNumber(int n) {
    if(n!=4) {
         System.out.print(n);
         printNumber(n + 1);
    }
}

you will find the output: 123

The reason for that is at each recursion call it will print the number n then it will call it another time for n+1 until it reach 4.

Upvotes: 0

khelwood
khelwood

Reputation: 59096

Each call to printNumber calls printNumber(n+1) first, and then prints something out. So printNumber(n+1) has already finished before printNumber(n) prints its number out.

main():
   printNumber(1):
       printNumber(2):
           printNumber(3):
               printNumber(4) // does nothing
               print(n) // "3"
               // printNumber(3) completes, returning control to printNumber(2)
           print(n) // "2"
           // printNumber(2) completes, returning control to printNumber(1)
       print(n) // "1"
       // printNumber(1) completes, returning control to main()
   // main() completes

Upvotes: 7

rghome
rghome

Reputation: 8819

The variable n is either 1 2 or 3.

What are you printing first, n or n + 1? You are printing n + 1 first, then n. So the numbers must come out in reverse order.

Upvotes: 1

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