[boost] Re: [format] conversion to string

From: Vladimir Prus (ghost_at_[hidden])
Date: 2005-03-31 02:20:51

Rob Stewart wrote:

>> 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();
> }

For format, such function already exists. However, it would be nice to have
it fully generic -- just a more conveniently spelled synonym for
lexical_cast. "lexical_cast" is just too long for a commonly used and
no-so-dangerouns (unlike, say reinterpret_cast), functionality.

> 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.

Or the other way around. The generic version will use stream io, and
specific version will call .str()

- Volodya