Determine Output void main() static int i=5 if(--i) main() printf( d...

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C Miscellaneous

Determine Output:

void main()
{
      static int i=5;
      if(--i){
            main();
            printf("%d ", i);
      }
}

A) 5 4 3 2 1 B) 0 0 0 0 C) Infinite Loop D) None of These

Answer & Solution

Correct Answer: Option B

The variable "i" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

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Nagaveni Varada 3 years ago

i is initially 5.
main() is called recursively.
i is decremented to 4, and main() is called recursively.
i is decremented to 3, and main() is called recursively.
i is decremented to 2, and main() is called recursively.
i is decremented to 1, and main() is called recursively.
i is decremented to 0, and main() is called recursively.
i is printed (0).
Control goes back to the previous recursive call. i is printed (1).
Control goes back to the previous recursive call. i is printed (2).
Control goes back to the previous recursive call. i is printed (3).
Control goes back to the previous recursive call. i is printed (4).
Therefore, the output of the program will be 0000.

Expired Director 6 years ago

I think it will be 4 3 2 1

Madhuri Anagha 8 years ago

But printf statement is inside if loop, so when the value of variable i is 0, control comes out of the loop and there end the program. So none of these is the correct answer option D .

Rahul Khatwani 9 years ago

Sir. In this program the print function is never called. As every till the i becomes 0 it returns to main() with executing the print function. And when the i becomes 0 the loop condition fails and it breaks.

Suraj Yadav 10 years ago

output will be:4 3 2 1

Sourav Nanda 10 years ago

plz explain
souravkumarnanda@gmail.com

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