Structure of a C++ Object in Memory Vs a Struct

Question

If I have a class as follows

   class Example_Class 
   {
       private:
         int x; 
         int y; 
       public: 
         Example_Class() 
         { 
             x = 8;
             y = 9;
         }
       ~Example_Class() 
       { } 
   };

And a struct as follows

struct
{
   int x;
   int y;
} example_struct;

Is the structure in memory of the example_struct simmilar to that in Example_Class

for example if I do the following

struct example_struct foo_struct;
Example_Class foo_class = Example_Class();

memcpy(&foo_struct, &foo_class, sizeof(foo_struct));

will foo_struct.x = 8 and foo_struct.y = 9 (ie: the same values as the x,y values in the foo_class) ?

The reason I ask is I have a C++ library (don't want to change it) that is sharing an object with C code and I want to use a struct to represent the object coming from the C++ library. I'm only interested in the attributes of the object.

I know the ideal situation would be to have Example_class wrap arround a common structure between the C and C++ code but it is not going to be easy to change the C++ library in use.

Answer 1

The C++ standard guarantees that memory layouts of a C struct and a C++ class (or struct -- same thing) will be identical, provided that the C++ class/struct fits the criteria of being POD ("Plain Old Data"). So what does POD mean?

A class or struct is POD if:

• All data members are public and themselves POD or fundamental types (but not reference or pointer-to-member types), or arrays of such
• It has no user-defined constructors, assignment operators or destructors
• It has no virtual functions
• It has no base classes

About the only "C++-isms" allowed are non-virtual member functions, static members and member functions.

Since your class has both a constructor and a destructor, it is formally speaking not of POD type, so the guarantee does not hold. (Although, as others have mentioned, in practice the two layouts are likely to be identical on any compiler that you try, so long as there are no virtual functions).

See section [26.7] of the C++ FAQ Lite for more details.