Adding number to pointer value

Question

I am trying to add a number to a pointer value with the following expression:

&AddressHelper::getInstance().GetBaseAddress() + 0x39EA0;

The value for the &AddressHelper::getInstance().GetBaseAddress() is always 0x00007ff851cd3c68 {140700810412032}

should I not get 0x00007ff851cd3c68 + 0x39EA0 = 7FF81350DB08 as a result?

while I am getting: 0x00007ff851ea3168 or sometimes 0x00007ff852933168 or some other numbers.

Did I took the pointer value incorrectly?

Pointer arithmetic take into account type of the pointer... — Jarod42
– Jarod42, Commented Oct 6, 2021 at 10:14
(0x00007ff851ea3168 - 0x00007ff851cd3c68) / 0x39EA0 = 0x08, So I suspect it is the size of your struct/type. — Jarod42
– Jarod42, Commented Oct 6, 2021 at 10:21
@Kaihaku: See Pointer_arithmetic. it is pedantically UB, even if it might works in practice for most compiler. — Jarod42
– Jarod42, Commented Oct 6, 2021 at 10:26
@Kaihaku The standard decides what is and isn’t UB. Saying that something is “UB on an obscure microcontroller” doesn’t make sense. — Konrad Rudolph
– Konrad Rudolph, Commented Oct 6, 2021 at 10:31
Pedantically, you cannot have/use arbitrary adresses. Then compiler might decide what to do for some UB, but it is no longer inside C++ rules. Platform might provide "buffer" at specific address. Notice also that some compilers remove code which leads to UB. — Jarod42
– Jarod42, Commented Oct 6, 2021 at 10:41

Jarod42 · Accepted Answer · 2021-10-06 10:51:59Z

3

With pointer arithmetic, type is taken into account,

so with:

int buffer[42];
char* start_c = reinterpret_cast<char*>(buffer);
int *start_i = buffer;

we have

start_i + 1 == &buffer[1]
reinterpret_cast<char*>(start_i + 1) == start_c + sizeof(int).
and (when sizeof(int) != 1) reinterpret_cast<char*>(start_i + 1) != start_c + 1

In your case:

0x00007ff851ea3168 - 0x00007ff851cd3c68) / 0x39EA0 = 0x08

and sizeof(DWORD) == 8.

answered Oct 6, 2021 at 10:29

Jarod42

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