Jeremiah Willcock <jewillco@osl.iu.edu> writes:

On Wed, 25 Jan 2012, Eric Niebler 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. Is there a reason that the use of decltype is not on by default for compilers that support it, perhaps with a flag to force emulation mode (like Boost.Move supports)?

It's not the default because when we made it the default, all heck broke loose. Partly it was because of buggy libraries that returned a different type from operator() than their nested result<> template declared. Partly, it was due to a bug in the specification of decltype that was only resolved in closing minutes with the adoption of N3276[*]. And partly it is because of continuing non-compliance of compilers that don't property implement N3276.

Would it be possible to use decltype to at least handle C++11 lambdas (which are the one case where normal result_of is broken)? Is there a type trait to detect those?

-- Jeremiah Willcock

As a template argument a lambda is indistinguishable from any other functor (it may also have a conversion to a function pointer), so I don't think this is possible. You can try to turn the feature on by defining the macro yourself (with Boost.Config) and check if anything breaks. Your users however will be in for nice surprise if they include your headers and get a C++11 result_of which breaks boost libraries that they might include themselves. -- Philipp Moeller

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 <typename This, typename... Args> struct result<This(Args...)> { typedef decltype(boost::declval<This>()(boost::declval<Args>()...)) type; }; // construction and assignment /* … */ // invoke // precondition: f_ is initialized template <typename... Args> typename result<Func(Args&&...)>::type operator()(Args&&... args) { return (*f_)(static_cast<Args&&>(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