[boost] [range] as_literal always returning iterator_range

Subject: [boost] [range] as_literal always returning iterator_range
From: Tobias Germer (tgermer_at_[hidden])
Date: 2011-05-03 08:22:55

Hi all,

I think there is a problem with the behavior of boost::as_literal. If I
got it right, it is supposed to create an iterator_range from
null-terminated strings, which is fine. However, it also creates
iterator_ranges from exisiting ranges, i.e., it converts general ranges
to iterator_ranges. I think as_literal should fall back to identity
(i.e., just return its argument) if its argument is already a range.

With the current behavior, the type information of the range is lost.
This is a problem if I write generic code and want to dispatch several
cases based on the type of the range. Consider the following example:

------------------------------- 
template<class Rng> void f( Rng const& r ) {...} 
template<class T> void f( iterator_range<T*> const& r ) {...}
void f( std::string const& s ) {...} 
void f( std::vector const& v ) {...}

template<class Rng>
void g( Rng const& r ) {
        f(as_literal(r));
        ...
}

-------------------------------

Function g() accepts general ranges as well as zero-terminated strings.
It dispatches its argument using the type of the range and the overloads
of f(), in order to execute special code for some ranges like strings
and vectors. (In my particular case, I have special treatment for ranges
of contiguous memory.) 

This is not possible anymore after calling as_literal, because it always
returns an iterator_range. Wouldn't it be better to return the original
range, for example like this:

template< class Rng >
typename enable_if< has_range_const_iterator<Rng>, 
Rng &>::type as_literal (Rng& r ) {
        return r;
}

Regards,
Tobias

--
Dr. Tobias Germer | tgermer_at_[hidden]
Software Engineer
think-cell Software GmbH | Chausseestr. 8/E | 10115 Berlin | Germany
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