Question

Trying to find a “simple to use” safe_bool idiom/implementation, I’ve ended up with my own.

Q: Is this implementation correct?

template <typename T>
class safe_bool 
{
protected:
   typedef void (safe_bool::*bool_type)() const;
   bool_type to_bool_type(bool b) const 
     { return b ? &safe_bool<T>::safe_bool_true : 0; }

private:
   void safe_bool_true() const {}

private:
   bool operator ==(safe_bool<T> const & rhs);
   bool operator !=(safe_bool<T> const & rhs);
};

to be used like this:

struct A : public safe_bool<A>
{
   // operator bool() const { return true; }
   operator bool_type() const { return to_bool_type(true); }
};

The only addition to existing base classes would be to_bool_type, but I hope I’ve got everything else correct, too.

The test cases I used (VC9) can be found here.

The downsides I see in implementation: bool_type and to_bool_type are visible in derived classes, which might not appease to everyone. Also, using the wrong template argument (e.g. class B : public safe_bool<A> introduced during copy-and-paste) will go unnoticed.

Answer 1

Using a pointer to a member function as the bool alias is idiomatic, as you’re doing here.

Your implementation looks correct for what is there, but slightly incomplete. See http://en.wikibooks.org/wiki/More_C%2B%2B_Idioms/Safe_bool

IMO safe_bool falls into the category of things which do more harm than good; ie the complexity and confusion introduced by this idiom along with the mental effort needed to understand it are greater than the intial problem it is intending to solve.