Question

I have a class like this:

class Object {
public: 
    unsigned char data[8];
    // other variables
    // functions etc...
 };

The question is – are the object members all stored in the same place in memory relative to the object? So if I have an array: Object array[3], given a char pointer
char* data_ptr = array[0].data, will
data_ptr + (sizeof(Object)) then always point to array[1].data?

(I’ve read a couple of Q/As about how there might be padding inbetween data members of classes and structs – but I don’t think they answer my question.)

Thanks in advance,
Ben

Answer 1

sizeof Object already includes all internal padding of the class Object. including any padding at its end. Arrays don’t allow any additional padding. Therefore it is true that data_ptr + sizeof Object will have the address of array[1].data.

However I’m not sure if this is actually allowed. That is, the compiler might be allowed to assume that you never add a value larger than 8 (the size of the member array data) to array[0].data, and therefore it might apply optimizations that fail if you violate the rules. That is, your code might actually exhibit undefined behavior (which is the standard term for “the compiler is allowed to do anything in this case”).

However since you are using a pointer to char, for which there are more permissive rules (you can do many things with char* which you could not do with general types), it may be that it’s actually defined behaviour anyway.