Question

Can someone please explain me this peculiar output:

#include <stdio.h>

typedef struct node
{
  int i;
  struct node *next;
}node;

main()
{
    node *p,*q;
    printf(" %u ",sizeof(node));              // 16
    p = (node *)malloc(sizeof ( node ) ) ;     
    printf(" %p ",p);                    // 0x1cea010
    q = (node *)malloc(sizeof ( node ) ) ; 
    printf("\n %p ",q);                    // 0x1cea030
}

I have a 64 bit processor. When the size is shown to be 16 byes, why is 32 byte allocated for the node??
I checked out a 32- bit machine. The addresses had a separation of 8 bytes. With no padding and stuff. So is the difference of 4 bytes solely cause of some padding issue of the 64 bit machine??

Answer 1

Two malloc calls aren’t necessarily going to return consecutive memory areas. A better way to do this test would be:

main()
{
    node *p;
    printf(" %u ",sizeof(node));
    p = (node*)malloc(2 * sizeof (node));     
    printf(" %p \n %p ", &p[0], &p[1]);
    free(p);
}

By allocating an array, you can be sure that they are back-to-back in memory.

Depending on your implementation of malloc, your system may be using the memory in between p and q to store bookkeeping information that is used by realloc, free, and friends.