I have a 64-bit i7 machine. Suppose I allocate memory for n 32-bit integers. How many physical registers will actually be used in the allocation: n, or n/2? I tried to write the following simple programme to find out. #include <iostream> #include <cstdlib> using namespace std; int main (int argc, char *argv[]) { int a[4]; cout << &a[0] << "\t" << &a[3] << endl; cin.ignore (1); return 0; } // End main () The output is: 0018FA04 0018FA10 They seem further apart than they should be. Why aren't the addresses 04 and 07? And does this mean that the system is actually allocating four (or more) integers, instead of packing the four 32-bit integers into two 64-bit registers? Thanks in advance for your help.