Question

I am wondering why the value of register integer a is not overwritten:

#include <stdio.h>
#include <conio.h>

main()
{
    register int a=10;
    {
        register int a=30;
    }
    printf("%d",a);
    getch();
}

Why isn’t the value of a overwritten? The output is showing 10.

Answer 1

Having those braces inside your function creates a new nested block. That means that your line:

register int a=30;

is declaring a new variable that also happens to be called a. If you add another call to printf, you’ll see that happening (I rewrote your program a bit to remove the unnecessary register keywords and cleaned up some things to make it be standard C, too):

#include <stdio.h>

int main(void)
{
    int a = 10;
    {
        int a = 30;
        printf("%d\n", a);
    }
    printf("%d\n", a);
    return 0;
  }

The output from this program is:

30
10

The inner a (the one set to 30) is said to shadow the other local variable. You can create an arbitrary number of nestings, if you want. For example, this next program:

#include <stdio.h>

int main(void)
{
    int a = 10;
    {
        int a = 30;
        {
            int a = 40;
            printf("%d\n", a);
        }
        printf("%d\n", a);
    }
    printf("%d\n", a);
    return 0;
}

Produces this output:

40
30
10

The redeclarations of a hide all other declarations of a outside the current scope/block. What all of this means is that your original program is semantically equivalent to:

#include <stdio.h>

int main(void)
{
    int a = 10;
    printf("%d\n", a);
    return 0;
}

Since the inner a was set but never used. In fact, turning on some warning flags might let the compiler give some diagnostic messages about what’s going on. I used clang here:

$ clang -Wunused-variable example.c -o example
example.c:7:9: warning: unused variable 'a' [-Wunused-variable]
    int a = 30;
        ^
1 warning generated.