Question

Given the following code:

class foo;

foo* instance = NULL;

class foo
{
public:
   explicit foo(int j)
    : i(j)
   {
      instance = this;
   }

   void inc()
   {
      ++i;
   }

private:
   int i;
};

Is the following using defined behavior?

const foo f(0);

int main()
{
   instance->inc();
}

I’m asking because I’m using a class registry, and as I don’t directly modify f it would be nice to make it const, but then later on f is modified indirectly by the registry.

EDIT: By defined behavior I mean: Is the object placed into some special memory location which can only be written to once? Read-only memory is out of the question, at least until constexpr of C++1x. Constant primitive types for instance, are (often) placed into read-only memory, and doing a const_cast on it may result in undefined behavior, for instance:

int main()
{
    const int i = 42;
    const_cast<int&>(i) = 0; // UB
}

Answer 1

Yes, it is undefined behavior, as per 7.1.5.1/4:

Except that any class member declared mutable (7.1.1) can be modified, any attempt to modify a const object during its lifetime (3.8) results in undefined behavior.

Note that object’s lifetime begins when the constructor call has completed (3.8/1).