c++ casting a union to one of its member types

Question

The following seems perfectly logical to me, but isn't valid c++. A union cannot be implicitly cast to one of it's member types. Anyone know a good reason why not?

union u {
  int i;
  char c;
}
function f(int i) {
}
int main() {
  u v;
  v.i = 6;
  f(v);
}

And can anyone suggest a clean alternative (the cleanest I can come up with is f(v.i);, which I admit is very clean, but the above just seems even cleaner)

Answer 1

While agreeing with Crazy Eddie that it doesn't look that better to me you can actually get an implicit conversion by defining it:

union u {
    int i;
    char c;
    operator int () const { return i; }
    operator char () const { return c; }
};

Answer 2

How would the compiler know which member to use? It would need to keep track of which member was last assigned so it knows what to convert to. This is called a tagged union, and while it's certainly possible for the language to specify such a thing, that's not the case (it's a hold-over from C).

But that's okay, because we have boost::variant. Not only is it safer and more flexible than a union, it's more powerful. You can apply visitors to the variant, and it'll call (visit) the specified function with whichever member is currently active, with no further work from the user.