Re: [boost] [format] conversion to string

From: Rob Stewart (stewart_at_[hidden])
Date: 2005-03-30 11:07:08

From: Vladimir Prus <ghost_at_[hidden]>
> 
> While it's generally considered a bad practice to provide implicit 
> conversions, I think it would be good to provide such a conversion for the  
> boost::format class. Currently, I can't write:
> 
>   string s;
>   s = format("%1%") % 10;
> 
> and need to write
> 
>   s = (format("%1%") % 10).str();
> 
> Maybe it's specifics of my code, but I can count 10 such conversions in a 
> single source file. I think that:
> 
> 1. Calling .str() is rather inconvenient.
> 2. The 'format' class most often used inside expressions. It's not likely to 
> be passed around to functions and so the dangers of implicit conversions are 
> not critical. 

While your usage might be proven normative and thus could be used
to justify adding an implicit conversion, how about another
approach?  Why not add a non-member function str() or
to_string():

template <class T>
std::string
boost::str(T const & source_i)
{
   return source_i.str();
}

Then, you example becomes:

   s = str(format("%1%") % 10);

(I don't even know the types involved in your expression to make
a non-templated function.  Besides, I thought this might be
generally useful.)

To generalize further (warning: uncompiled, untested code):

template <class T>
typename boost::result_of<&T::str()>::type
boost::str(T const & source_i)
{
   return source_i.str();
}

That would be more flexible.  This could also be the primary
template specialization and it could be specialized for other
types that don't provide a str() member function.

-- 
Rob Stewart                           stewart_at_[hidden]
Software Engineer                     http://www.sig.com
Susquehanna International Group, LLP  using std::disclaimer;