Is a pointer the same as a byte address?

Question

Assume I have an array declared like this:

int *p = new int[size];

And p will of course point to the address of the first element. But since a byte is the smallest addressable unit of memory, does p actually point to the first byte of the first 4 bytes of the first element of the array?

Answer 1

The address of an int is not necessarily exactly the same as the address of the first byte ( char ) in its object representation. This is because some machines have native pointer registers that lack bits, such that sizeof (char *) != sizeof (int *) is possible.

Discounting that, though, the int * is convertible to the pointer to the first byte of the object representation via static_cast< char * >( p ) . You can pass the resulting pointer to std::memcpy to initialize another int or any POD class type whose first member is an int . ("First byte" therefore is defined as the one with the lowest address.)

For any machine you're likely to encounter in general-purpose computing, char * and int * are physically the same thing; their differences are just enforced by the compiler for the purpose of code safety. But there do exist exotic architectures where static_cast does something meaningful in this situation and something like reinterpret_cast would entirely fail perform the correct conversion.

Answer 2

Yes, it indeed points to the first of the four bytes. In the case of big Endian, the four bytes will look like 0 0 0 1. In the case of little Endian, the bytes will look like 1 0 0 0. So, just to be sure, in the case of big Endian, the first byte will contain a 0 while in the case of little Endian, the first byte will contain a 1.

Answer 3

In short, no. Because p is initialized as an int*, each element of p will take the number of bytes that an int does. To clarify:

int* p = new int[4];
for(int i = 0; i < 4; i++)
    p[i] = i;
cout << p[0] << " " << p[1] << " " << p[2] << " " << p[3];

This would output "0 1 2 3." You do not have to worry about individual bytes. In general, this is also true:

*p = p[0];