这个C ++代码是否可移植？

Question

struct A {
  int a;
  virtual void print() {}
};

struct B : A {
  int b;
  virtual void print() {}
};

int main(void)
{
  A *a = new B;
  a[0].print(); // g++, vs2010, clang call B::print.
}

All three g++, vs2010, clang call B::print. Almost doubting my C++ 101. I was under the impression that using a dot with an object does not result in a dynamic dispatch. Only -> with a pointer and dot with a reference will result in a dynamic dispatch.

So my question is whether this code is portable?

Answer 1

a[0] is the same as *a , and that expression is a reference to an A , and virtual dispatch happens through references just as it does through pointers. a->print() and (*a).print() are entirely identical.

Answer 2

It's portable. a[0] returns a reference, and references use dynamic dispatch as well.

Answer 3

Yes. It is equivalent to -

a->print();

Answer 4

It is portable and the behaviour is well-defined. operator[] on a pointer just does pointer arithmetic and a dereference. It is adding 0 * sizeof(A) to the pointer, so in a sense it is a "dangerous" operation as any other value but 0 would fail (on an array of Bs), but as 0 * sizeof(A) is 0, in this case you're ok because it's adding 0.

Polymorphism works on references as well as pointers.

Answer 5

Using a[0] with a pointer is well defined and means the same as *(a + 0) . That's how the built in [] operator works.

You are partly right about the compiler not needing to use dynamic dispatch when there is no polymorphism. This is just a common optimization though, and not required by the language.

When the code is

A a;
a.print();

the compiler can call the correct function directly, because it can tell the object type at compile time.