Re: [boost] [type_traits][function_types] Discard param const qualification, bug or feature?

On Mon, 30 Sep 2013 05:24:45 -0700, Rob Stewart wrote:

On Sep 29, 2013, at 11:00 PM, Mostafa wrote:

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On Sun, 29 Sep 2013 14:44:52 -0700, Sergey Zhuravlev wrote:

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Signature type int(int, const std::string) can be used as parameter for some compiletime algorithm. For example, algorithm that generate new signature type with optimal transfer arguments int (int, const std::string&) or generate signature with all argument references int (int&, const std::string&)

That was exactly the use case and the problem I had. Fortunately for me, I was able to specify the return type and parameters_type (as an mpl::vector) separately, which solved the issue. But I don't know if this is doable in general. For example, what does one do in the following scenario:

template <typename T> struct Foo { void mybar( // Construct efficient argument transfer signature for T::bar ) { T::bar(....); } };

without forcing the user to seperately specify T::bar paramtypes as a separate template parameter?

#include ... boost::call_traits<typename T::bar>::param_type

I should have been more specific, bar is some member function of T. So in Foo:mybar, the goal is to reconstruct T::bar's paramtypes as "efficient types". That's what the other person's post was also referring to. So if a client passes the following struct as a template parameter to Foo: struct ClientClass { static void bar(int const) { ... } }; the library is able to instantiate the following Foo::mybar void Foo::mybar(int const & x) { ClientClass::bar(x); }