Why It Gives Different Address with Same Variable?

Here's the output of your program on my system:

-12115289320x7ffeb7c9891c

The two outputs are jammed together on one line, making it difficult to tell which is which.

Adding newlines to the output, I get:

344225740
0x7ffc148477cc

(There's no guarantee that a variable will have the same address from one run of a program to the next, and some systems deliberately change memory allocations.)

The %d format specifier requires an argument of type int. The correct format specifier for a pointer value is %p, and it requires an argument of type void*. Using inconsistent types, as you've done in your program, causes undefined behavior. At best, on many systems pointers are 64 bits and int is 32 bits, so %d can't possibly show an entire pointer value. It might show part of the pointer value, it might show garbage, or it might crash.

The %p format typically uses hexadecimal, but the format is implementation-defined.

Here's a corrected version of your program:

#include <stdio.h> 
int main(void)
{
    int a = 5;
    int *ptra = &a;
    printf("%p\n", (void*)&a);
    printf("%p\n", (void*)ptra);
}

and the output on my system:

0x7ffd62a733ec
0x7ffd62a733ec

You'll see different output, but the two lines should match each other.

Note: void main() is incorrect; the correct declaration is int main(void).

They are the same number. You're printing one in decimal form, the other in hexadecimal form.

#include<stdio.h> 
int main(int argc, char **argv)
{
    int a=5;
    int *ptra=&a;
    printf("%p\r\n",&a);
    printf("%p\r\n",ptra);
    return 0;
}

prints

0x7ffc8e4cb974
0x7ffc8e4cb974

for me

This happens due to the difference in format strings.

%d    takes 32 bits and displays it as a signed value  123
%p    takes a pointer and display it in address format    0fef:0004

Actual value is same but there is difference of interpreting the value.

As already mentioned in the answer from Keith Thompson () the reason is that you are printing an integer value by using %d but the argument isn't an integer - it is a pointer value. Providing an argument with a type that doesn't match the format specifier (i.e. providing a pointer value for %d) leads to undefined behavior which means that it may print any value, text or even nothing.

All that is already well covered by Keith Thompson ()

I just like to add that in case you actually want to print a decimal integer representation of the pointer value, you can use the types intptr_t or uintptr_t. Like:

#include <stdio.h> 
#include <inttypes.h> 
int main(void)
{
    int a = 5;
    int *ptra = &a;
    printf("%p\n", (void*)&a);
    printf("%p\n", (void*)ptra);
    
    intptr_t dptr = (intptr_t)(void*)ptra;
    uintptr_t uptr = (uintptr_t)(void*)ptra;

    printf("A signed integer representation of the same pointer in decimal        : %" PRIdPTR "\n", dptr);
    printf("An unsigned integer representation of the same pointer in decimal     : %" PRIuPTR "\n", uptr);
    printf("An unsigned integer representation of the same pointer in hexadecimal : %" PRIxPTR "\n", uptr);

}

Possible output:

0x7ffcaceba19c
0x7ffcaceba19c
A signed integer representation of the same pointer in decimal        : 140723209609628
An unsigned integer representation of the same pointer in decimal     : 140723209609628
An unsigned integer representation of the same pointer in hexadecimal : 7ffcaceba19c

Using %d escape sequence in your first printf is converting your reference from hexadecimal to decimal, i.e, (x)₁₆ --> (x)₁₀.

So your code should be:

#include<stdio.h>

int main(void)
{
    int a=5;
    int *ptra=&a;
    printf("%p\n",&a);
    printf("%p",ptra);
    return 0;
 }

Output: (may vary on your machine)

000000000061fe14
000000000061fe14