Re: [Boost-Users] What's safe_bool?

22 Nov 2001

      From: 
...
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 
// ===>
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.