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为什么行为如此不同？#if 1版本成功（这很奇怪）编译并产生预期的输出到stdout，但版本#if 0没有：

#include <iostream>
#include <type_traits>

#include <cstdlib>

template< typename T >
class X
{

    struct B {};

public :

    struct U : B {};

};

template< typename T >
struct Y
{

    using X_type = X< T >;

    template< typename D,
              typename = typename std::enable_if< std::is_base_of< typename X_type::B, D >::value >::type >
    void operator () (D const &) const
    {
        std::cout << "allowed only for derived from B" << std::endl;
    }

};

template< typename T >
struct Z
{

    using X_type = X< T >;

    template< typename D >
    auto
    operator () (D const &) const
    -> typename std::enable_if< std::is_base_of< typename X_type::B, D >::value >::type
    {
        std::cout << "allowed only for derived from B!" << std::endl;
    }

};

int main()
{
    using T = struct W; // not matters
    using X_type = X< T >;
    using U = typename X_type::U;
#if 0
    using V = Y< T >;
#else
    using V = Z< T >;
#endif
    V v;
    v(U());
    return EXIT_SUCCESS;
}

这会产生一个错误（根据我对这两种情况的期望（实际上），本质上“B”是私有的）：

<stdin>: In function 'int main()':
<stdin>:60:10: error: no match for call to '(V {aka Z<main()::W>}) (U)'
<stdin>:34:8: note: candidate is:
<stdin>:41:5: note: template<class D> typename std::enable_if<std::is_base_of<typename X<T>::B, D>::value>::type Z<T>::operator()(const D&) const [with D = D; T = main()::W]
<stdin>:41:5: note:   template argument deduction/substitution failed:
<stdin>: In substitution of 'template<class D> typename std::enable_if<std::is_base_of<typename X<T>::B, D>::value>::type Z<T>::operator()(const D&) const [with D = D; T = main()::W] [with D = X<main()::W>::U]':
<stdin>:60:10:   required from here
<stdin>:10:12: error: 'struct X<main()::W>::B' is private
<stdin>:41:5: error: within this context

g++ -v输出包含以下行：

gcc version 4.8.1 (rev3, Built by MinGW-builds project)

我希望B除X.

该问题仅在类模板中异常发生，而对于纯类则没有（这里我们有两个相同的问题，因为两个变体都不必编译，但它们都已编译）。

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