Re: What's safe_bool?====Thanks!!!

From: mr_lilirong_at_[hidden]
Date: 2001-11-22 20:57:33

Thank you very much. Now I understand it.

--- In Boost-Users_at_y..., "Peter Dimov" <pdimov_at_m...> wrote:
> From: <mr_lilirong_at_y...>
> > The following code is excerpted from Boost.Function. I just don't
> > unstanding it. Why use "safe_bool"? What's the advantage? Can any 
one
> > explain it to me?
> >
> > Thanks.
> >
> > // code from <boost/function/function_base.hpp>
> > // ===>
> >
> >   private:
> >     struct dummy {
> >       void nonnull() {};
> >     };
> >
> >     typedef void (dummy::*safe_bool)();
> >
> >   public:
> >     operator safe_bool () const
> >       { return (this->empty())? 0 : &dummy::nonnull; }
> 
> safe_bool, as its name implies, is a safer type to use in implicit
> conversions when you need your type to act as a bool, like in
> 
> if(f)
> 
> Using an operator bool() - since a bool can be converted to int - 
means that
> any arithmetic expression that contains 'f' will compile, like
> 
> f == 5
> f + 2
> f >> 1
> 
> etc
> 
> Using 'operator void const * () const', as std::cin and std::cout 
do, is a
> bit better, but it still allows some unfortunate side effects:
> 
> if(std::cin == std::cout) delete std::cin;
> 
> Using a member pointer avoids most of these pitfalls leaving only
> comparisons valid. They are specifically disabled in 
boost::function.
> 
> --
> Peter Dimov
> Multi Media Ltd.