C++11 嵌套 std::conditional

Question

这个表达式有什么问题吗？

template<class T, class FieldT>
using addRefU = typename std::conditional<
                            // ok when true. When false result has no reference at all
                            false,     
                            typename std::add_rvalue_reference< FieldT >::type,
                            typename std::conditional<
                                true,
                                typename std::add_rvalue_reference< FieldT >::type,
                                typename std::add_lvalue_reference< FieldT >::type
                            >
                        >::type;

int main()
{
        std::cout << std::is_rvalue_reference<
            addRefU<A, B>
        >::value << std::endl;
        std::cout << std::is_lvalue_reference<
            addRefU<A, B>
        >::value << std::endl;

}

http://coliru.stacked-crooked.com/a/21593805f2c6e634

因此，它根本没有引用。是否不允许嵌套 std::conditional？

Answer 1

您在嵌套的 conditional 上忘记了 ::type:

template<class T, class FieldT>
using addRefU = typename std::conditional<
                            // ok when true. When false result has no reference at all
                            false,     
                            typename std::add_rvalue_reference< FieldT >::type,
                            typename std::conditional<
                                true,
                                typename std::add_rvalue_reference< FieldT >::type,
                                typename std::add_lvalue_reference< FieldT >::type
                            >::type
                        >::type;

int main()
{
        std::cout << std::is_rvalue_reference<
            addRefU<A, B>
        >::value << std::endl;
        std::cout << std::is_lvalue_reference<
            addRefU<A, B>
        >::value << std::endl;

}

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