What is the output of the program?void main(){ static int i=5; if(--i){ main () printf ("%d",i);}}A4,3,2,1,0BCompilation ErrorC5,4,3,2,1D0,0,0,0

Break Down the Problem

1. Analyze the recursive main() function and understand the flow of execution.
2. Identify the role of the static variable i and how it is decremented in each function call.
3. Assess the output of the printf function.

Relevant Concepts

1. Static Variables: The variable i is declared static, meaning it retains its value between function calls.
2. Recursive Function Calls: The function calls itself while decrementing the value of i.
3. Output Printing: The output is determined by the execution of printf after the recursive calls.

Analysis and Detail

1. Initially, i is set to 5.
2. In the first call to main(), --i is evaluated:
   • i becomes 4, and the condition is true (non-zero), so main() is called again.
3. In the second call:
   • i becomes 3 (decremented), and main() is called again.
4. In the third call:
   • i becomes 2, and main() is called again.
5. In the fourth call:
   • i becomes 1, and main() is called again.
6. In the fifth call:
   • i becomes 0. The condition if(--i) is false, so main() doesn't call itself further.
7. After returning to the previous call (where i was 1), printf("%d", i); executes, which outputs 0.
8. As the calls unwind, the next outputs will be done in reverse order of calls:
   • For i = 1, it outputs 0
   • For i = 2, it outputs 1
   • For i = 3, it outputs 2
   • For i = 4, it outputs 3

Verify and Summarize

The execution of the recursive function prints the value of i in reverse order of how the function was called. The output from the function is thus 0 followed by 1, 2, 3, 4.

Final Answer

The output of the program is "43210" (Option A).