使用SFINAE select接口来实现

Question

我正在尝试 select 具体 class 基于模板参数实现的接口。在下面的简化示例中，有两个接口具有不同名称的方法 - Implementation class 需要提供方法实现，具体取决于模板参数。

如果我使用 IntType - 它应该实现 IntInterface（getInt() 和 setInt(int)）。如果我使用 DoubleType - 它应该实现 DoubleInterface （getDouble() 和 setDouble(double)）。

为了实现这一点，我创建了一个特征class，它决定了应该使用哪个接口（InterfaceType），但也有另一个参数用于SFINAE。

我的想法是，如果我使用例如MyTypeTrait<Type>::MyTypeInt 但相关特征 class 没有 MyTypeInt 定义，编译器将抛出这种可能的重载（例如 setInt()），并使用另一个。这就是假人应该发挥作用的地方——他们有不同的论点，而且他们不是虚拟的。

但是，它不起作用。请参阅下面的编译器错误。

我正在使用 Visual Studio 2013 (VC12)。

struct IntInterface
{
    virtual int getInt() const = 0;
    virtual void setInt(int value) = 0;
};

struct DoubleInterface
{
    virtual double getDouble() const = 0;
    virtual void setDouble(double value) = 0;
};

const int IntType = 0;
const int DoubleType = 1;

template <int Type>
struct MyTypeTrait;

template <>
struct MyTypeTrait<IntType>
{
    using MyTypeInt = int;
    using InterfaceType = IntInterface;
};

template <>
struct MyTypeTrait<DoubleType>
{
    using MyTypeDouble = double;
    using InterfaceType = DoubleInterface;
};

template <int Type>
struct Implementation : public MyTypeTrait<Type>::InterfaceType
{
    // Actual interface implementation for the case of IntType
    virtual typename MyTypeTrait<Type>::MyTypeInt getInt() const override { return 0; }
    virtual void setInt(typename MyTypeTrait<Type>::MyTypeInt value) override {}

    // Dummys for SFINAE - to be used in the case of DoubleType
    typename int getInt(int) const { return 0; }
    void setInt() {}

    // Actual interface implementation for the case of DoubleType
    virtual typename MyTypeTrait<Type>::MyTypeDouble getDouble() const override { return 0.0; }
    virtual void setDouble(typename MyTypeTrait<Type>::MyTypeDouble value) override {}

    // Dummys for SFINAE - to be used in the case of IntType
    typename double getDouble(int) const { return 0.0; }
    void setDouble() {}
};


int main(int argc, char* argv[])
{
    Implementation<IntType> myInt;
    Implementation<DoubleType> myDouble;
}

编译器错误：

1>c++-tests.cpp(50): error C2039: 'MyTypeDouble' : is not a member of 'MyTypeTrait<0>'
1>          c++-tests.cpp(26) : see declaration of 'MyTypeTrait<0>'
1>          c++-tests.cpp(61) : see reference to class template instantiation 'Implementation<0>' being compiled
1>c++-tests.cpp(50): error C2146: syntax error : missing ';' before identifier 'getDouble'
1>c++-tests.cpp(50): error C2433: 'Implementation<0>::MyTypeDouble' : 'virtual' not permitted on data declarations
1>c++-tests.cpp(50): error C4430: missing type specifier - int assumed. Note: C++ does not support default-int
1>c++-tests.cpp(50): warning C4183: 'getDouble': missing return type; assumed to be a member function returning 'int'
1>c++-tests.cpp(51): error C2039: 'MyTypeDouble' : is not a member of 'MyTypeTrait<0>'
1>          c++-tests.cpp(26) : see declaration of 'MyTypeTrait<0>'
1>c++-tests.cpp(51): error C2061: syntax error : identifier 'MyTypeDouble'
1>c++-tests.cpp(55): error C2535: 'void Implementation<0>::setDouble(void)' : member function already defined or declared
1>          c++-tests.cpp(51) : see declaration of 'Implementation<0>::setDouble'
1>c++-tests.cpp(50): error C3668: 'Implementation<0>::getDouble' : method with override specifier 'override' did not override any base class methods
1>c++-tests.cpp(51): error C3668: 'Implementation<0>::setDouble' : method with override specifier 'override' did not override any base class methods
1>c++-tests.cpp(42): error C2039: 'MyTypeInt' : is not a member of 'MyTypeTrait<1>'
1>          c++-tests.cpp(33) : see declaration of 'MyTypeTrait<1>'
1>          c++-tests.cpp(62) : see reference to class template instantiation 'Implementation<1>' being compiled
1>c++-tests.cpp(42): error C2146: syntax error : missing ';' before identifier 'getInt'
1>c++-tests.cpp(42): error C2433: 'Implementation<1>::MyTypeInt' : 'virtual' not permitted on data declarations
1>c++-tests.cpp(42): error C4430: missing type specifier - int assumed. Note: C++ does not support default-int
1>c++-tests.cpp(42): warning C4183: 'getInt': missing return type; assumed to be a member function returning 'int'
1>c++-tests.cpp(43): error C2039: 'MyTypeInt' : is not a member of 'MyTypeTrait<1>'
1>          c++-tests.cpp(33) : see declaration of 'MyTypeTrait<1>'
1>c++-tests.cpp(43): error C2061: syntax error : identifier 'MyTypeInt'
1>c++-tests.cpp(47): error C2535: 'void Implementation<1>::setInt(void)' : member function already defined or declared
1>          c++-tests.cpp(43) : see declaration of 'Implementation<1>::setInt'
1>c++-tests.cpp(42): error C3668: 'Implementation<1>::getInt' : method with override specifier 'override' did not override any base class methods
1>c++-tests.cpp(43): error C3668: 'Implementation<1>::setInt' : method with override specifier 'override' did not override any base class methods

Answer 1

对于您是想支持 either/or 接口还是这 2 个接口的 complete/NOP 版本的问题，我不是很确定。

这是前者的一种解决方案。

#include <utility>
#include <type_traits>

struct IntInterface
{
    virtual int getInt() const = 0;
    virtual void setInt(int value) = 0;
};

template<class TrueFalse>
struct IntInterfaceImpl {};

template<>
struct IntInterfaceImpl<std::true_type> : IntInterface
{
    int getInt() const override { return i_; }
    void setInt(int value) override { i_ = value; }

    int i_;
};

struct DoubleInterface
{
    virtual double getDouble() const = 0;
    virtual void setDouble(double value) = 0;
};

template<class TrueFalse>
struct DoubleInterfaceImpl {};

template<>
struct DoubleInterfaceImpl<std::true_type> : DoubleInterface
{
    double getDouble() const override { return i_; }
    void setDouble(double value) override { i_ = value; }

    double i_;
};


enum type {
    is_int,
    is_double
};

template<type T>
struct Implementation
: IntInterfaceImpl<std::integral_constant<bool, T == is_int>>
, DoubleInterfaceImpl<std::integral_constant<bool, T == is_double>>
{

};


int main()
{
    Implementation<type::is_int> i {};
    i.setInt(6);
    int a = i.getInt();

    Implementation<type::is_double> d {};
    d.setDouble(6.0);
    int b = d.getDouble();
}

Answer 2

这个：

// Actual interface implementation for the case of IntType
virtual typename MyTypeTrait<Type>::MyTypeInt getInt() const override { return 0; }
virtual void setInt(typename MyTypeTrait<Type>::MyTypeInt value) override {}

// Dummys for SFINAE - to be used in the case of DoubleType
int getInt(int) const { return 0; }
void setInt() {}

工作无望。 SFINAE 仅适用于模板实例化的直接上下文中的替换失败。这些成员函数不在 class 模板实例化的直接上下文中，因此 MyTypeTrait<DoubleType>::MyTypeInt 是一个 硬错误 ，而不是替换失败。

您将不得不重新考虑您的设计。可能，您只有两个完全独立的接口实现，然后编写如下内容：

template <int Type>
using Implementation = std::conditional_t<
    Type == IntType,
    IntImplementation,
    DoubleImplementation>;

或者只是对所有这些进行明确的模板专业化。

使用SFINAE select接口来实现

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