C Programming - Functions - Discussion
#include<stdio.h>
int reverse(int);
int main()
{
int no=5;
reverse(no);
return 0;
}
int reverse(int no)
{
if(no == 0)
return 0;
else
printf("%d,", no);
reverse (no--);
}
Step 1: int no=5; The variable no is declared as integer type and initialized to 5.
Step 2: reverse(no); becomes reverse(5); It calls the function reverse() with '5' as parameter.
The function reverse accept an integer number 5 and it returns '0'(zero) if(5 == 0) if the given number is '0'(zero) or else printf("%d,", no); it prints that number 5 and calls the function reverse(5);.
The function runs infinetely because the there is a post-decrement operator is used. It will not decrease the value of 'n' before calling the reverse() function. So, it calls reverse(5) infinitely.
Note: If we use pre-decrement operator like reverse(--n), then the output will be 5, 4, 3, 2, 1. Because before calling the function, it decrements the value of 'n'.
Is it right ?. If Yes/No then why?
I also agree with both of you.
Is the answer is correct ?
If yes then please explain it.
When reverse (no) is called recursively, it will every time get the value as 5 and not the decremented value as 4, 3, 2 and so on. This is because every time reverse (no) is called, the variable named "no" is a new different variable not the same. The decremented value is in the previous function call (or stack frame) and therefore after calling the reverse (no) we are entering a new stack frame resulting into a new variable "no" which is having a value 5 every time. Hence the result will be infinite loop.
Hope this helped.