Where is the variable definition in this simple line of code? - c

Question

I'm a little confused about this:
#include <stdio.h>
#define MAXLINE 1000
int x;
int main()
{
int x;
getch();
return 0;
}
Where is the variable definition in this code? I'm assuming that it would be the external variable. In that case, shouldn't the variable in the function have an extern modifier?
What if the external variable was below the main function?

Answer 1

Example 1:
int x; // declares and defines global variable
int main()
{
int x; // declares and defines *new* local variable, which hides (shadows) the global variable **in this scope**
}
Example 2:
int main()
{
extern int x; // declares variable that will refer to variable defined *somewhere*
}
int x;
Example 3:
int x; // declares and defines global variable
int main()
{
extern int x; // redundant, declares variable that will refer to variable defined *somewhere*, but it is already visible in this scope
}

Answer 2

extern doesn't mean outside the current scope, it means an object with external linkage. An automatic variable never has external linkage, so your declaration int x inside main can't possibly refer to that. Hence it's hiding the global int x, that is, the the variable x with auto storage class is hiding the global x. You need to read more about storage classes in C
Refer the below program AFTER reading about them :
#include <stdio.h>
int i = 6;
int main()
{
int i = 4;
printf("%d\n", i); /* prints 4 */
{
extern int i; /* this i is now "current". */
printf("%d\n", i); /* prints 6 */
{
int *x = &i; /* Save the address of the "old" i,
* before making a new one. */
int i = 32; /* one more i. Becomes the "current" i.*/
printf("%d\n", i); /* prints 32 */
printf("%d\n", *x); /* prints 6 - "old" i through a pointer.*/
}
/* The "previous" i goes out of scope.
* That extern one is "current" again. */
printf("%d\n", i); /* prints 6 again */
}
/* That extern i goes out of scope.
* The only remaining i is now "current". */
printf("%d\n", i); /* prints 4 again */
return 0;
}