我是否正确地排除了 base 的 fn() 函数？

Question

我正试图做一些有趣的事情，但它在 VC++ 2015. I tried it in clang and g++ 上崩溃了，没有编译或运行时错误。（链接指向 rextester.com 演示）

#include <type_traits>
#include <iostream>

int fn1()    { return 1; }
int fn2(int) { return 2; }

template <typename FN_T, FN_T FN, typename DERIVED>
class test_base
{
    template <typename T = DERIVED, typename = std::enable_if_t<!T::fn_specified>>
    int fn()
    {
        return FN();
    }

public:
    int fn_indirect()
    {
        return static_cast<DERIVED*>(this)->fn();
    }
};

template <typename FN_T, FN_T FN, typename ENABLER = void>
class test_derived : public test_base<FN_T, FN, test_derived<FN_T, FN>>
{
public:
    static constexpr bool fn_specified = false;
};

template <typename FN_T, FN_T FN>
class test_derived<FN_T, FN, std::enable_if_t<FN == &fn2>>
    : public test_base<FN_T, FN, test_derived<FN_T, FN>>
{
    using base = test_base<FN_T, FN, test_derived<FN_T, FN>>;
    friend base;

public:
    static constexpr bool fn_specified = true;
    int fn()
    {
        return FN(1);
    }
};

int main(void)
{
    test_derived<decltype(&fn1), &fn1> x1;
    test_derived<decltype(&fn2), &fn2> x2;

    std::cout << x1.fn_indirect() << " ";

    // comment next line out and it'll work in VC++
    std::cout << x2.fn_indirect() << std::endl;
    return 0;
}

这个想法是，如果我在派生的 class 中指定一个 fn()，那么基础 class 将不会实例化该函数，因此不会由于以下原因而导致错误调用 FN() 时参数过多，当 FN() 指向特定函数时（在本例中为 fn2(int)）。

我这样做是否正确，还是我遗漏了什么？

Answer 1

您使用的结构称为 Expression SFINAE。 Visual Studio 直到 2017 年才声明为 support it. Visual Studio 2015's updates enables partial support (read more about update 1, updates 1, 2 and 3)。然而，要使其在所有版本的 MSVC 中工作，需要创建不涉及表达式的变通方法，例如内部模板专业化参数。要在您的情况下实现它，可以使用带有附加间接层的模式匹配（例如 std::integral_constant）：

#include <type_traits>
#include <iostream>

int fn1()    { return 1; }
int fn2(int) { return 2; }

template <typename FN_T, FN_T FN, typename DERIVED>
class test_base
{
    template <typename T = DERIVED, typename = std::enable_if_t<!T::fn_specified>>
    int fn()
    {
        return FN();
    }

public:
    int fn_indirect()
    {
        return static_cast<DERIVED*>(this)->fn();
    }
};

template <class FN_T>
class test_derived : public test_base<typename FN_T::value_type, FN_T::value, test_derived<FN_T>>
{
public:
    static constexpr bool fn_specified = false;
};

template <typename FN_T>
class test_derived<std::integral_constant<FN_T, &fn2>>
    : public test_base<FN_T, &fn2, test_derived<std::integral_constant<FN_T, &fn2>>>
{
    using base = test_base<FN_T, &fn2, test_derived<std::integral_constant<FN_T, &fn2>>>;
    friend base;

public:
    static constexpr bool fn_specified = true;
    int fn()
    {
        return fn2(1);
    }
};

int main(void)
{
    test_derived<std::integral_constant<decltype(&fn1), &fn1>> x1;
    test_derived<std::integral_constant<decltype(&fn2), &fn2>> x2;

    std::cout << x1.fn_indirect() << " ";

    // comment next line out and it will work
    std::cout << x2.fn_indirect() << std::endl;
    return 0;
}

[live demo]

我是否正确地排除了 base 的 fn() 函数？

Am I correctly excluding the base's fn() function?

c++

visual-c++

c++14

sfinae

crtp