Memory allocation in C and C++ arrays

Array indices are just a convenient and more meaningful way of doing addition.
Hence the first element in any array, static or dynamic, is 0 as you add 0 to find it.

char *szDynamicStr = new char[12];
strcpy(szDynamicStr, "hello world");
printf("%c == %c\n", szDynamicStr[0], //This accesses the first element ('h') as an array offset
	*(szDynamicStr + 0)); //This accesses the first element as an addition to the pointer value

Both szDynamicStr[0] and *(szDynamicStr + 0) compile to exactly the same code, because the array translates to the pointer offset.

The pointer is pointing to the first element in the array, so offset 0 is the first element, offset 1 is the 2nd element, etc...

If this answers your question and you don't care why it works stop reading here or you might get confused.

When adding a number to a pointer, the number is multiplied by the size of the data that is getting pointed to (in this case 1 byte for a single ASCII char) by the compiler.

This is better explained in code:

DWORD pNumbers[] = { 0xD00D, 0xBADF00D }; //This is still a pointer, it is just easier to allocate this example this way.
printf("%u == %u\n", pNumbers[1], //The 2nd element in the array (0xBADF00D) as you would expect
	*(pNumbers + 1)); //This is also the 2nd element in the array. This actually adds 1 * sizeof(DWORD) = 4 bytes to the pointer in order to get the 2nd element in the array.