Determine Output:
void main()
{
static int i=5;
if(--i){
main();
printf("%d ", i);
}
}
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: Option B
Solution(By Examveda Team)
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.This Question Belongs to C Program >> C Miscellaneous
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Comments ( 5 )
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Determine Output:
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I think it will be 4 3 2 1
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 .
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.
output will be:4 3 2 1
plz explain
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