Question

Show the output of the following code: public class Test1 { public static void main(String[] args) { System.out.println(f2(2, 0)); } public static int f2(int n, int result) { if (n == 0) return 0; else return f2(n - 1, n + result); } }

Answer 1

0

Step-by-step explanation:

Given the code as follows:

1. public class Test1 {
2. public static void main(String[] args) {
3. System.out.println(f2(2, 0));
4. }
6. public static int f2(int n, int result)
7. {
8. if (n == 0)
9. return 0;
10. else
11. return f2(n - 1, n + result);
12. }
13. }

The code above shows a recursive function calling (Line 11). The f2 function will call itself when n is not equal to 0. In the main program, f2 function is called by passing argument 2 and 0 to parameter n and result, respectively (Line 3).

Since the n is not equal to 0, and therefore else statement will run (Line 11). f2 function is called recursively for several rounds:

Round 1 - f2(2-1, 2 + 0) - > f2 (1, 2)

Round 2 - f2(1 -1 , 1 + 2) -> f2(0, 3)

f2(0 , 3) will return 0 and the 0 will be returned to the calling point at f2(1, 2). At the end, the same 0 value will be return from the f2 function.