Thread: Program show result without extern

I am wondering how program is showing correct output without external storage external class in source files because I have two source file

main.c

Code:

    #include <stdio.h>  
  void fun ();
  
  int main ()
  
  {
	  int x = 1;
	  
	  fun (x);
	  
      return 0;
  }

FileA.c

Code:

  #include <stdio.h>
  void fun ( int  y)
  {
	  y++;
	  
	  printf ("x = %d \n", y);
	  
  }

output : 2

variable x is declared in main.c source file and this variable is being passed to function fun which is in source file FileA.c

Originally Posted by Salem

Because your function prototype on line 2 of main.c is extern by default.

I have declared variable x in main.c file. I want to access this variable in another file FileA.c. That's why i used extern storage class

Code:

#include <stdio.h> 
  extern int x;  

int main ()

{
     x = 1;

    fun (x);

    return 0;
}

FileA.c

Code:

 

  #include <stdio.h>
  
 void fun ( )
  {
	  x++;
	  
	  printf ("x = %d \n", x);
	  
  }

I don't know why compiler show error error: 'x' undeclared (first use in this function)

Originally Posted by Salem

Read all your other 'extern' threads again!
extern function
extern variable is not working as expected

I am surprised because I am accessing a variable x that is declared in another source file file1.c without extern

main.c

Code:

 #include <stdio.h>
#include "file1.h"

int main() {
  x = 6;
  printf("%d\n", x);

  return 0;
}

header

Code:

int x;

file1.c

Code:

int x;

output : 6
Does compiler makes variable as extern variable by default ?