Storage management in C

Question

I wrote the program to find out about memory management in C. But I noticed something strange about this program: Why is the variable "a" in memory location 6487580 and not in memory location 6487576.

Because if a pointer is 8 bytes, then it should only reach up to this address. And then the variable "a" would have space there.

#include <stdio.h>
#include <stdlib.h>

int main(){
    int a=10;
    int *ptr;
    ptr=&a;
    int **ptrptr;
    ptrptr=&ptr;

    printf("adress:\n");
    printf("Adress a: %i\n",&a);
    printf("Sizeof a: %i\n",sizeof(a));
    printf("Adress ptr: %i\n",&ptr);
    printf("Sizeof ptr: %i\n",sizeof(ptr));
    printf("Adress ptrptr: %i\n",&ptrptr);
    printf("Sizeof ptrptr: %i\n",sizeof(ptrptr));

    return 0;
}

Answer 1

Memory alignment. It is likely that the computer you're running on can retrieve data from memory faster or more efficiently if it's stored on a doubleword (8 byte) boundary, or at least the compiler is programmed to make that assumption. If you examine the addresses in binary you'll note they all end in 00, putting them on a doubleword boundary.

Thus, ptrptr is stored on a doubleword boundary at the low-order location and takes 8 bytes. ptr is stored next in memory, and occupies the next 8 bytes. Then 4 bytes are skipped (6487576-6487579) so that a will be aligned on a doubleword boundary at 6487580.

Your compiler may have flags which instruct it to align variables on word or doubleword boundaries - or, conversely, to ignore such alignment concerns. Consult your local documentation for such information.