Re: [Boost-users] BOOST_RESULT_OF_USE_DECLTYPE not default on C++11 compilers

27 Jan 2012

      Jeremiah Willcock wrote:
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On 1/25/2012 6:42 PM, Jeremiah Willcock wrote:
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At least on GCC 4.6 in C++0x (C++11) mode, BOOST_RESULT_OF_USE_DECLTYPE
is not set by default, preventing classes such as
boost::transform_iterator from working on built-in lambda expressions.
<snip>
So far I'm using a -D on the compile line to turn it on;
BOOST_RESULT_OF_USE_DECLTYPE might break some codes of Boost,
so how about wrapping a function object without tr1 result_of support into
a tr1 function object (i.e. a function object with tr1 result_of support)?
This approach enables us to use transform_iterator with C++11 lambda
without defining BOOST_RESULT_OF_USE_DECLTYPE.

I just wrote a code which wraps into a tr1 function object. Additionally,
I made wrapped functors to be DefaultConstructible and CopyAssignable.
(Since C++11 lambda is not DefaultConstructible nor CopyAssignable,
transform_iterator with it is not a "decent" iterator. Like oven::regular,
the wrapper solves this problem.)

Here is a code:
(Complete code is attached in this mail)

template <typename Func>
class tr1_functor
{
public:
    // tr1 result_of support
    template <typename Sig>
    struct result;

    template 
    struct result
    {
        typedef
decltype(boost::declval<This>()(boost::declval<Args>()...)) type;
    };

    // construction and assignment
	/* … */

    // invoke
    // precondition: f_ is initialized
    template 
    typename result::type operator()(Args&&... args)
    {
        return (*f_)(static_cast(args)...);
    }

    // const version of operator()
    /* … */

private:
    // data
    boost::optional<Func> f_;
};

template <typename Func>
tr1_functor<Func> make_tr1_functor(Func const& f)
{
    return tr1_functor<Func>(f);
}

/* Usage */
int main (int argc, char* argv[])
{
    using boost::adaptors::transformed;

    BOOST_FOREACH(int i
      , boost::irange(0,5)
      | transformed(make_tr1_functor([](int i){ return i * 100; }))
    )
    {
        std::cout << i << " ";
    }
    // Output: "0 100 200 300 400"

    return 0;
}

Regards,
Michel

Re: [Boost-users] BOOST_RESULT_OF_USE_DECLTYPE not default on C++11 compilers

Michel Morin