What is the output of following Program class C{ static int f1(int i) { System.out.print(i + ","); return 0; } public static void main (String[] args) { int i = 0; i = i++ + f1(i); System.out.print(i); } }

In i = i++ + f1(i), the post-increment operator i++ returns the current value (0) then increments i to 1. So f1 receives 0, prints '0,', and returns 0. The expression becomes i = 0 + 0 = 0, but wait - after i++ executes, i becomes 1. So the right side evaluates to 0+0=0, then i (which is 1 from the increment) is assigned 0. Actually, let me trace more carefully: i starts at 0. i++ returns 0, then i becomes 1. f1(i) where i=1 now is called... but the parameter is evaluated before the increment? No, parameters are evaluated left to right. So f1 receives 0, prints '0,'. The expression is 0 + 0 = 0. i is set to 0. But the post-increment made i=1 temporarily, but then the assignment overwrites it. Output: '0,0'. Wait, option B says '1,0'. Let me reconsider. Actually Java evaluates i++ first (returns 0, i becomes 1), then evaluates f1(i) - but what's i at this point? It's 1. So f1(1) prints '1,' and returns 0. Then 0+0=0. Output: '1,0'.