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

"shallow" versus "deep" copy is less meaningful in C++ than it is in C or Java.

To illustrate this, I've changed your Foo class from three ints to an int, an int*, and a vector:

#include 
#include 

class Foo {
public:
  int a;
  int *b;
  std::vector c;
};

using namespace std;

int main() {
  Foo f1, f2;
  f1.a = 42;
  f1.b = new int(42);
  f1.c.push_back(42);
  f2 = f1;

  cout << "f1.b: " << f1.b << " &f1.c[0]: " << &f1.c[0] << endl;
  cout << "f2.b: " << f2.b << " &f2.c[0]: " << &f2.c[0] << endl;
}

When this program is run, it yields the following output:

f1.b: 0x100100080 &f1.c[0]: 0x100100090
f2.b: 0x100100080 &f2.c[0]: 0x1001000a0

The int is boring, so I've left it out. But look at the difference between the int* and the vector: the int* is the same in f1 and f2; it's what you would call a "shallow copy". The vector however is different between f1 and f2; it's what you would call a "deep copy".

What's actually happened here is that the default operator = in C++ behaves as if the operator = for all of its members were called in order. The operator = for ints, int*s, and other primitive types is just a byte-wise shallow copy. The operator = for vector performs a deep copy.

So I would say the answer to the question is, No, the default assignment operator in C++ does not perform a shallow copy. But it also doesn't perform a deep copy. The default assignment operator in C++ recursively applies the assignment operators of the class's members.