Question

Given a function such as:

template< typename T >
void function1( const T &t )
{
  function2( boost::lexical_cast<std::string>(t) );
}

What kind of overhead is incurred if the type passed to function1 is already a std::string?

Does the overhead vary, depending on the type I’m lexical_cast-ing to?

Is it superfluous to make an overloaded function to bypass the cast? E.g.:

void function1( const std::string &t )
{
  function2( t );
}

template< typename T >
void function1( const T &t )
{
  function1( boost::lexical_cast<std::string>(t) );
}

The version of boost may be relevent to your answer, as I understand that lexical_cast has received a few optimizations across revisions.

Answer 1

Since the documentation doesn’t offer anything on this topic, I dug into the lexical_cast source (1.51.0) and found that it does some compile-time checking on the types and decides a specific “caster class” that does the conversion. In case source and target are the same, this “caster class” will simply return the input.

Pseudo-codified and simplified from source (boost/lexical_cast.hpp:2268):

template <typename Target, typename Source>
Target lexical_cast(const Source &arg)
{
    static if( is_character_type_to_character_type<Target, src> ||
               is_char_array_to_stdstring<Target, src> ||
               is_same_and_stdstring<Target, src> )
    //         ^-- optimization for std::string to std::string and similar stuff
    {
      return arg;
    }
    else
    {
      /* some complicated stuff */
    }
}

I can’t directly see any optimizations for other identity casts, though, and looking through the normally selected lexical_cast_do_cast “caster class” is making my head hurt. 🙁