Question

is it possible to return the sizeof a derived class already from base class/struct?

imho the size of a class is a kind of property of itself, like the weight of a human being. But I don’t want to write the same function in every class.

many thanks in advance
Oops
PS: so code to make my question more clear:

template <typename T>
struct StructBase {
    size_t size<T>(){
        return sizeof(T);
    }
};

struct MyStruct: public StructBase<MyStruct> {
    double d1;
    double d2;
    double d3;
    MyStruct(): d1(0), d2(0), d3(0){}
    //I do not want to do this here
    //size_t size(){ //return size<MyStruct> ;}
};

int main(void) {
   MyStruct m;
   std::cout << m.size(); //nop ?!?
}

Answer 1

You can’t overload the sizeof operator. And I’m not sure what you are asking:

class A {
   int x, y;
};

then:

size_t n = sizeof( A );

gives you the size of class A – I don’t see where derivation comes into this.

Edit: and regarding your expanded question, I don’t see why you can’t say:

MyStruct m;
std::cout << sizeof( m );

Note that size is not a property like weight – weight can change, and is different for different instances. The size of a class must be known by the compiler at compile time and cannot ever change.