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'.
reverse(no--);===>calling processing.
reverse(no);
no=no-1;
So because in that situation before decrement the we called the reverse function in recursive..
Hence the result will be infinite loop.
No chance get decrement.
reverse(--no);==>no=no-1; after that call reverse(no);
So print : 5, 4, 3, 2, 1.
no++ or ++no no effect output infinite.
Can you please tell me why we get same output ie infinite in case of post increment and pre increment?
Reverse function will repeats itself until no becomes zero.
I think output must be 5, 4, 3, 2, 1.
And when no becomes zero, function will exit from loop.
Here, no-- means
no=no-1; ->no=5 is assigned first on "LHS" and then "RHS" no gets decremented. so the final value we get to return is 5 only which is assigned to "LHS". And the same value is again called in reverse() function.
So the same procedure is repeated infinite time.
Hope this will help you
Why zero is not included please explain why we avoid zero?