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'.
if(x>0)myfunc(--x);
printf("%d",x);
}
int main()
{
myfunc(5);
return 0;
}
output: 001234
Anyone, please explain this.
Why zero is not included please explain why we avoid zero?
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