Re: [Boost-Users] Re: shared_ptr - implicit conversion to bool

From: "David Abrahams"

"wasekvesely " writes:

...

I've found this in "shared_ptr.hpp":

// implicit conversion to "bool" typedef T * (this_type::*unspecified_bool_type)() const;

operator unspecified_bool_type() const // never throws { return px == 0? 0: &this_type::get; }

Why this is better than:

operator bool() const { return px != 0; }

What more can I do with somethin like this?

It's not that you can do more; quite the opposite.

int f(long); int x = f(shared_ptr<int>()); // error

With operator bool(), that (and many other abuses) would compile.

Yeah. It's the same reason that stream classes has an implicit conversion to pointer, to give a bool value regarding its state. With an implicit conversion to bool, you could have done nonsensical things like the above example, and e.g.: int value=cin+cin; // Would compile! Instead, using the current definition, you may do: delete cin; // (!) This is fixed in shared_ptr, though, so you can't do that, either. Regards, Terje