Recursion using the main function

Question

I need to find the output of this code. I know that this is recursion but i do not exactly know how this code actually works. I thought that this would be an infinite code but it is not. Can someone please explain me?

 #include <stdio.h>
 char *str = "kre";
 void main(void) {
     int i;
     static int j = 3;
     for (i = 0; i < j; i++) {
         printf("%d ", j--);
         main();
     }
     printf("%sn%c ", str, 'i');
 }

suggestion: single-step through your program with a debugger. — pmg, May 22 '20 at 07:51
suggestion 2: use an [indentation style](https://en.m.wikipedia.org/wiki/Indentation_style) other than "everything left-aligned" — pmg, May 22 '20 at 07:52

score 0 · Accepted Answer · answered May 22 '20 at 08:08

0

I thought that this would be an infinite code but it is not

It's not an infinite code because, you declare static for j (see What does “static” mean in C?):

static int j = 3;

The value of j will be updated when you call main function. You can see the output of your program:

3 2 1 kreni kreni kreni kreni

the value of j decrement from 3 to 1, then when j = 0, you can not access the for loop because the condition i < j is not accepted, so the main() is not called again.

If you want an infinite code, you remove static from declaration of j, your program will never be stopped.

answered May 22 '20 at 08:08

Hitokiri

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Thanks for explanation, but i still do not understand why it prints "kreni" 4 times when the loop is finished. It is maybe a stupid question but this really confuses me. – Petar May 22 '20 at 08:19
@Petar because, in the last call of `main()`, `j=0`, so you cannot access the `for` loop to print the value of `j` (`printf("%d ", j--);` is called 3 times), but when `j=0` you can still print `kreni`. this is reason why you print `j` 3 times but 4 times for `kreni` – Hitokiri May 22 '20 at 08:23
I thought that output would be "3 kreni 2 kreni 1 kreni" and it confuses me how it prints "3 2 1" in the loop but then prints "kreni" 4 times. – Petar May 22 '20 at 08:38
The function is called when `i=0`, `i=1` and `i=2`. However, don't forget that the first time `main` is called by starting the program. – Cheatah May 22 '20 at 08:42
@Petar each time you call main, your program back to the first line of main, until `j = 0`, 4 called times of `main` will can reach the line `printf("%sn%c ", str, 'i');`. You can print the value of `j` in the last `printf`: `printf("%sn%c, j = %d ", str, 'i',j);` to see – Hitokiri May 22 '20 at 08:46
Thanks for your answers. I understand how it works, but I think that when it prints "3 2 1" it still does not reach to the last line, so I think the output should be "3 2 1 kreni" because the code would reach to the last printf only when it can not enter the for loop. – Petar May 22 '20 at 09:00
@Petar `main` is called 4 times (1 in starting program, then 3 in for loop). So `printf("%sn%c ", str, 'i');` is reached in 4 times. – Hitokiri May 22 '20 at 09:04
@Petar Why would it not reach the last line? Which condition would stop the program from continuing after the `main` call or exiting the `for` loop? – Cheatah May 22 '20 at 09:05
@Cheatah Yes, do attention with `i = 0` in `for` loop. The value of `i` maybe indeterminate before `for` loop, but at starting of `for` loop, `i = 0` – Hitokiri May 22 '20 at 09:06

Recursion using the main function

1 Answers1