如何确定编译器是否在虚函数上使用早期绑定或后期绑定？

Question

I have the following code:

class Pet {
public:
  virtual string speak() const { return ""; }
};

class Dog : public Pet {
public:
  string speak() const { return "Bark!"; }
};

int main() {
  Dog ralph;
  Pet* p1 = &ralph;
  Pet& p2 = ralph;
  Pet p3;

  // Late binding for both:
  cout << "p1->speak() = " << p1->speak() <<endl;
  cout << "p2.speak() = " << p2.speak() << endl;

  // Early binding (probably):
  cout << "p3.speak() = " << p3.speak() << endl;
}

I have been asked to determine whether the compiler uses early or late binding for the final function call. I have searched online but have found nothing to help me. Can someone tell me how I would carry out this task?

Answer 1

You can look at the disassembly, to see whether it appears to be redirecting through a vtable.

The clue is whether it calls directly to the address of the function (early binding) or calls a computed address (late binding). The other possibility is that the function is inlined, which you can consider to be early binding.

Of course the standard doesn't dictate the implementation details, there may be other possibilities, but that covers "normal" implementations.

Answer 2

You can always use hack :D

//...
Pet p3;
memset(&p3, 0, sizeof(p3));
//...

If compiler does use vtbl pointer, guess what will gonna happen :>

p3.speak()  // here

Answer 3

Look at the generated code. Eg in Visual Studio you can set a breakpoint, then right-click and select "Go To Disassembly".

Answer 4

It uses early binding. You have an object of type P3. While it is a base class with a virtual function definition, the type is concrete and known at compile time, so it doesn't have to consider the virtual function mapping to derived classes.

This is much the same as if you called speak() in the Pet constructor - even when making derived objects, when the base class constructor is executing the type of the object is that of the base so the function would not use the v-table, it would call the base type's version.

Basically, early binding is compile time binding and late binding is run-time binding. Run time binding is only used in instances where the compiler doesn't have enough type information at compile time to resolve the call.

Answer 5

In fact the compiler has no obligation to use either one particularly, just to make sure that the right function is called. In this case, your object is of the concrete type Pet , so as long as Pet::speak is called the compiler is "doing the right thing".

Now, given that the compiler can statically see the type of the object, I suspect that most compilers will optimize away the virtual call but there is no requirement that they do so.

If you want to know what your particular compiler is doing the only way is to consult its documentation, source code, or the generated disassembly.

Answer 6

I just thought of a way to tell at runtime, without guesswork. You can simply override the vptr of your polymorphic classes with 0 and see if the method is called or if you get a segmentation fault. This is what I get for my example:

Concrete: Base
Concrete: Derived
Pointer: Base
Pointer: Derived
DELETING VPTR!
Concrete: Base
Concrete: Derived
Segmentation fault

Where Concrete: T means that calling the virtual member function of T through a concrete type was successful. Analogously, Pointer: T says that calling the member function of T through a Base pointer was successful.

---

For reference, this is my test program:

#include <iostream>
#include <string.h>

struct Base {
  unsigned x;
  Base() : x(0xEFBEADDEu) {
  }
  virtual void foo() const {
    std::cout << "Base" << std::endl;
  }
};

struct Derived : Base {
  unsigned y;
  Derived() : Base(), y(0xEFCDAB89u) {
  }
  void foo() const {
    std::cout << "Derived" << std::endl;
  }
};

template <typename T>
void dump(T* p) {
  for (unsigned i = 0; i < sizeof(T); i++) {
    std::cout << std::hex << (unsigned)(reinterpret_cast<unsigned char*>(p)[i]);
  }
  std::cout << std::endl;
}

void callfoo(Base* b) {
  b->foo();
}

int main() {
  Base b;
  Derived d;
  dump(&b);
  dump(&d);
  std::cout << "Concrete: ";
  b.foo();
  std::cout << "Concrete: ";
  d.foo();
  std::cout << "Pointer: ";
  callfoo(&b);
  std::cout << "Pointer: ";
  callfoo(&d);
  std::cout << "DELETING VPTR!" << std::endl;
  memset(&b,0,6);
  memset(&d,0,6);
  std::cout << "Concrete: ";
  b.foo();
  std::cout << "Concrete: ";
  d.foo();
  std::cout << "Pointer: ";
  callfoo(&b);
  std::cout << "Pointer: ";
  callfoo(&d);
  return 0;
}