Default assignment operator= in c++ is a shallow copy?

Question

Just a simple quick question which I couldn\'t find a solid answer to anywhere else. Is the default operator= just a shallow copy of all the class\' members on the right ha

Answer 1

I personally like the explanation in Accelerated c++. The text (page 201) says:

If the class author does not specify the assignment operator, the compiler synthesizes a default version. The default version is defined to operate recursively - assigning each data element according to the appropriate rules for the type of that element. Each member of a class type is assigned by calling that member's assignment operator. Members that are of built-in type are assigned by assigning their values.

As the members in the question are integers, I understand that they are deep copied. The following adaptation of your example illustrates this:

#include
class foo {
public:
  int a, b, c;
};

int main() {
    foo f1, f2;
    f1 = f2;
    std::cout << "f1.a and f2.a are: " << f1.a << " and " << f2.a << std::endl;
    f2.a = 0;
    std::cout << "now, f1.a and f2.a are: " << f1.a << " and " << f2.a << std::endl;
}

which prints:

f1.a and f2.a are: 21861 and 21861
now, f1.a and f2.a are: 21861 and 0