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?
1.Pre-increment/pre decrement.
2.Substitute value.
3.Evaluate.
4.Post Increment/post decrement.
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?
Can you please tell me why we get same output ie infinite in case of post increment and pre increment?
I also agree with both of you.
Is the answer is correct ?
If yes then please explain it.