what is the output??

Question

#include<stdio.h>
int i=5;
int main()
{
  extern int j;
  printf("\ni=%d \nj=%d",i,j);
  int j=10;
  return 0;
}


OUTPUT:-


i=5
j=10

but when i compile this code ,i get compilation error 

source of question is http://cprogrammingcodes.blogspot.in/2012/02/external-storage-class.html

please explain what,s wrong

Answer 1

Output and explanation in the link you provided are wrong!

• extern declaration specifies that the variable j is defined somewhere else.  
• The compiler passes the external variable to be resolved by the linker. 
• So compiler doesn't find any error.
• During linking the linker searches for the definition of j.
• Since it is not found the linker flags an error

​#include<stdio.h>

int i=5;
int main()
{
  extern int j;
  printf("\ni=%d \nj=%d",i,j);
 
  return 0;
}

 int j=10;  // define j outside main() block.! So linker successfuly finds definition and prints value!