If a pointer's address is 64 bits, why does it appear as 6 bytes when printed?

Question

#include <stdio.h>
int main(void){
  int *ptr;
  printf("the value of ptr is %p",ptr);
}

This gives me 0x7fffbd8ce900, which is only 6 bytes. Should it be 8 bytes (64bit)?

Answer 1

Although a pointer is 64 bits, current processors actually only support 48 bits, so the upper two bytes of an address are always either 0000 or (due to sign-extension) FFFF.

In the future, if 48 bits is no longer enough, new processors can add support for 56-bit or 64-bit virtual addresses, and existing programs will be able to utilize the additional space since they're already using 64-bit pointers.