LNK2019 with pure virtual assignment(=) operator in multilevel class hierarchy?

Question

Ok, maybe i'm too fatigued to think about the solution to this problem, I looked a lot about a similar problem on the internet but didn't find one. So here's my terrible code:

class X
{
public:
    virtual const X& operator =( const X& ) = 0;
};

class Y : public X
{
public:
         virtual const X& operator =( const X& x )
         {
             return *this;
         }
};

class Z : public Y
{
};


int main_08(int argc, char* argv[])
{
    Z z1;
    Z z2;
    z1 = z2;
    return 0;
}

According to my expectations, code should run fine because since the assignment of class Z is not overridden, it should point to the assignment operator of Y which is defined. So when writing "z1 = z2", the assignment operator of its base class should be called. The code runs fine when I comment out this line

Instead, I get LNK2019 saying:

---

error LNK2019: unresolved external symbol "public: virtual class X const & __thiscall X::operator=(class X const &)" (??4X@@UAEABV0@ABV0@@Z) referenced in function "public: class Y & __thiscall Y::operator=(class Y const &)" (??4Y@@QAEAAV0@ABV0@@Z)

---

I am puzzled and can't figure out how virtual function routing mechanism has made me call X::operator =( const X& ). Any thoughts?

Answer 1

Y doesn't have a copy assignment operator. It has an assignment operator from X, but that's not the same thing. So the compiler generates one. The generated operator automatically invokes X's assignment operator in a non-virtual manner, i.e. it actually wants to call X::operator=, not whatever overrides it.

Z also gets a generated assignment operator, which calls Y's assignment operator. And so you get a reference to the actual implementation of X::operator=, which you don't provide.

More fundamentally, though, assignment and hierarchies don't mix well. You will run into the slicing problem. Don't make classes that are part of a polymorphic hierarchy assignable (or copyable at all). It's just not a good idea.

Answer 2

The compiler-generated operator = for class Z will call the one in the base class.