在没有Boost的情况下实现BOOST_DEDUCED_TYPENAME

Question

有以下代码段:

template<typename ValueType>
ValueType any_cast(any & operand)
{
    typedef BOOST_DEDUCED_TYPENAME remove_reference<ValueType>::type nonref;


    nonref * result = any_cast<nonref>(&operand);
    if(!result)
        boost::throw_exception(bad_any_cast());

    // Attempt to avoid construction of a temporary object in cases when 
    // `ValueType` is not a reference. Example:
    // `static_cast<std::string>(*result);` 
    // which is equal to `std::string(*result);`
    typedef BOOST_DEDUCED_TYPENAME boost::mpl::if_<
        boost::is_reference<ValueType>,
        ValueType,
        BOOST_DEDUCED_TYPENAME boost::add_reference<ValueType>::type
    >::type ref_type;

    return static_cast<ref_type>(*result);
}

是否有可能实现BOOST_DEDUCED_TYPENAME不Boost？我只能使用C++11.

Answer 1

如果你有一个支持C++ 11的编译器,我发现你不太可能需要BOOST_DEDUCED_TYPENAME.即,它位于include/boost/config/suffix.hpp:

// BOOST_DEDUCED_TYPENAME workaround ------------------------------------------//
//
// Some compilers don't support the use of `typename' for dependent
// types in deduced contexts, e.g.
//
//     template <class T> void f(T, typename T::type);
//                                  ^^^^^^^^
// Replace these declarations with:
//
//     template <class T> void f(T, BOOST_DEDUCED_TYPENAME T::type);

#ifndef BOOST_NO_DEDUCED_TYPENAME
#  define BOOST_DEDUCED_TYPENAME typename
#else
#  define BOOST_DEDUCED_TYPENAME
#endif

只需使用typename关键字.