What main(++i) will return in C

Question

I have a program like this.

‪#include<stdio.h>
#include<stdlib.h>

int main(int i) {                /* i will start from value 1 */
if(i<10)                        
printf("\n%d",main(++i));        /* printing the values until i becomes 9 */    
}

output : 
5
2
2
2

Can anyone explain how the output is coming ?? what main(++i) is returning for each iteration. Also it is producing output 5111 if i remove the \n in the printf function. Thanks in advance.

Answer 1

First of all, the declaration of main() is supposed to be int main(int argc, char **argv). You cannot modify that. Even if your code compiles, the system will call main() the way it is supposed to be called, with the first parameter being the number of parameters of your program (1 if no parameter is given). There is no guarantee it will always be 1. If you run your program with additional parameters, this number will increase.

Second, your printf() is attempting to print the return value of main(++i), howover, your main() simply don't return anything at all. You have to give your function a return value if you expect to see any coherence here.

And finally, you are not supposed to call your own program's entrypoint, much less play with recursion with it. Create a separate function for this stuff.

Answer 2

Here's what the C Draft Standard (N1570) says about main:

5.1.2.2.1 Program startup

1 The function called at program startup is named main. The implementation declares no prototype for this function. It shall be defined with a return type of int and with no parameters:

int main(void) { /* ... */ }

or with two parameters (referred to here as argc and argv, though any names may be used, as they are local to the function in which they are declared):

int main(int argc, char *argv[]) { /* ... */ }

or equivalent or in some other implementation-defined manner.

Clearly, the main function in your program is neither of the above forms. Unless your platform supports the form you are using, your program is exhibiting undefined behavior.

Answer 3

This program has undefined behavior (UB) all over the place, and if you have a single instance undefined behavior in your program, you can't safely assume any output or behavior of your program - It legally can happen anything (although in real world the effects often are somewhat localized near the place of UB in the code. The old C90 standard listed are more than 100 (if i recall right) situations of UB and there is a not known number of UBs on top, which is behavior for situations, the standard do not describe. A set of situations, that are UB exists, for every C and C++ Standard.

In your case (without consulting standards) instances of UB are at least:

• not returning an value of a function that is declared with a return value. (exception: calling main the FIRST time - thanks, Jim for the comments)
• defining (and calling) main other than with the predefined forms of the standard, or as specified (as implementation defined behavior) by your compiler.

Since you have at least one instance of UB in your program, speculations about the results, are somewhat... speculative and must make assumptions about your compiler, your operating system, hardware, and even software running on parallel, that are normally not documented or can be known.

Answer 4

You are not initializing i, so by default value will be taken from the address where it is stored in RAM.

This code will produce garbage output if you run the code multiple times after restarting your computer.

The output will also depend on compiler.

Answer 5

I'm surprised that even compiles.

When the operating system actually runs the program and main() gets called, two 32 (or 64) bit values are passed to it. You can either ignore them by declaring main(void), or use them by declaring main(int argc, char** args).

As the above prototype suggests, the first value passed is a count of the number of command-line arguments that are being passed to the process, and the second is a pointer to where a list of these arguments is stored in memory, likely on the program's local stack.

The value of argc is always at least 1, because the first item string in args is always the name of the program itself, generated by the OS.

Regarding your unexpected output, I'd say something is not getting pulled off or pushed onto the stack, so variables are getting mixed up. This is either due to the incomplete argument list for main() or the fact that you've declared main to return an int, but haven't returned anything.

Answer 6

I think, the main method is calling itself inside the main method. to increment the value of a variable, i++ is printing the value of i before it increment while ++i it increment the value of i first before it print the value of i.

you could use this..

int x=0;
main()
{
   do
   {
   printf(x++);
   }while (i<10);

}