Question

情境：
我有多种类型可以归类为序列容器所有序列容器都是数据结构，但并非每个数据结构都是序列容器。

以下是代码中说明的示例。此示例中涉及的唯一“重要类型”是Array_T。它分为两类：它是一个序列容器，由于所有序列容器都是数据结构，因此它又是一个数据结构。

//A sequence container type
class Array_T{};

//A type trait for that particular sequence container
template <typename T> struct Is_Array           { static const bool value = false; };
template <>           struct Is_Array<Array_T>  { static const bool value = true;  };

//A type trait to identify all of the sequence containers
template <typename T> struct Is_A_Sequence_Container { static const bool value = Is_Array<T>::value
/* would probably "or" together more sequence types, but we only have Array_T in this example */;};

//A type trait to identify all of the data structures
template <typename T> struct Is_A_Data_Structure { static const bool value = Is_A_Sequence_Container<T>::value
/* would probably "or" together more data structure types, but we only have sequence containers in this example */;};

请注意，不能在Array_T上进行继承;它必须保持其声明的方式。

问题：
我想写两个函数。一个函数将处理所有序列容器，另一个函数将处理所有数据结构。我不知道序列容器函数是否真的存在，因为代码的那部分可能会生成也可能不生成。

那么，我如何使用元模板编程，为类型选择最接近的匹配标识？以下是预期行为的两个示例：

案例1：

// ...
//Both functions exist! Call the more specific one.
// ...

function(Array_T{}); // prints "sequence container"

案例2：

// ...
//Only the data structure one exists(not the sequence container one)
// ...

function(Array_T{}); // prints "data structure"

到目前为止我的尝试：

#include <iostream>
#include <type_traits>

//A sequence container type
class Array_T{};

//A type trait for that particular sequence container
template <typename T> struct Is_Array           { static const bool value = false; };
template <>           struct Is_Array<Array_T>  { static const bool value = true;  };

//A type trait to identify all of the sequence containers
template <typename T> struct Is_A_Sequence_Container { static const bool value = Is_Array<T>::value
/* would probably "or" together more sequence types, but we only have Array_T in this example */;};

//A type trait to identify all of the data structures
template <typename T> struct Is_A_Data_Structure { static const bool value = Is_A_Sequence_Container<T>::value
/* would probably "or" together more data structure types, but we only have sequence containers in this example */;};

// ↑ all of this code was already shown to you


//NOTE: This function MAY OR MAY NOT actually appear in the source code
//This function handles all sequence types
template<class T, typename std::enable_if<Is_A_Sequence_Container<T>::value,int>::type=0>
void function(T t) {
    std::cout << "sequence container" << std::endl;
    return;
}

//This function handles all data structures; assuming a more specific function does not exist(*cough* the one above it)
template<class T, typename std::enable_if<Is_A_Data_Structure<T>::value,int>::type=0>
void function(T t) {
    std::cout << "data structure" << std::endl;
    return;
}

int main(){

    function(Array_T{});
}

现在我意识到这不起作用，因为对于enable_ifs的两个值都是真的。
所以我想在数据结构函数中添加第二个enable_if来检查序列容器函数是否存在。像这样：

//...
//This function handles all data structures; assuming a more specific function does not exist(*cough* the one above it)
template<class T, typename std::enable_if<Is_A_Data_Structure<T>::value,int>::type=0,
                  typename std::enable_if</*if the more specific function does not exist*/,int>::type=0>>
void function(T t) {
    std::cout << "data structure" << std::endl;
    return;
}

int main(){

    function(Array_T{});
}

这就是我被困住的地方。有没有办法在不触及Array_T减速的情况下执行此操作，并且不涉及调度的第三个函数？

Answer 1

我会使用Tag调度：

struct DataStructureTag {};
struct SequenceContainerTag : public DataStructureTag {};

template <typename T> struct DataStructureTagDispatcher
{
    typedef typename std::conditional<Is_A_Sequence_Container<T>::value,
                                      SequenceContainerTag,
                                      DataStructureTag>::type type;
};


// NOTE: This function MAY OR MAY NOT actually appear in the source code
// This function handles all sequence types
template<class T>
void function(T&& t, const SequenceContainerTag&) {
    std::cout << "sequence container" << std::endl;
    return;
}

// This function handles all data structures (not handled my a more specific function)
template<class T>
void function(T&& t, const DataStructureTag&) {
    std::cout << "data structure" << std::endl;
    return;
}

template <class T>
typename std::enable_if<Is_A_Data_Structure<T>::value, void>::type
function(T&& t)
{
    typedef typename DataStructureTagDispatcher<T>::type tag;
    function(t, tag());
}

Answer 2

您还可以使用类层次结构来消除歧义的歧义

struct R2 {};
struct R1 : R2 {};

//NOTE: This function MAY OR MAY NOT actually appear in the source code
//This function handles all sequence types
template<class T, typename std::enable_if<Is_A_Sequence_Container<T>::value,int>::type=0>
void function(R1, T t) {
    std::cout << "sequence container" << std::endl;
    return;
}

//This function handles all data structures; assuming a more specific function does not exist(*cough* the one above it)
template<class T, typename std::enable_if<Is_A_Data_Structure<T>::value,int>::type=0>
void function(R2, T t) {
    std::cout << "data structure" << std::endl;
    return;
}

int main(){
    function(R1{}, Array_T{});
}

Answer 3

我改编了https://stackoverflow.com/a/264088/2684539：

// template funcName should exist
#define HAS_TEMPLATED_FUNC(traitsName, funcName, Prototype)                          \
    template<typename U>                                                             \
    class traitsName                                                                 \
    {                                                                                \
        typedef std::uint8_t yes;                                                    \
        typedef std::uint16_t no;                                                    \
        template <typename T, T> struct type_check;                                  \
        template <typename T = U> static yes &chk(type_check<Prototype, &funcName>*); \
        template <typename > static no &chk(...);                                    \
    public:                                                                          \
        static bool const value = sizeof(chk<U>(0)) == sizeof(yes);                  \
    }

然后

//NOTE: This function MAY OR MAY NOT actually appear in the source code
//This function handles all sequence types
template<class T, typename std::enable_if<Is_A_Sequence_Container<T>::value,int>::type = 0>
void function(T t) {
    std::cout << "sequence container" << std::endl;
    return;
}

// this assumes that any template 'function' exists
// (Do you have version for `data structure` ?)
// or else create a dummy struct and then
// template <T>
// typename std::enable_if<std::is_same<T, dummy>::value>::type function(dummy) {}
HAS_TEMPLATED_FUNC(isFunctionExist_Specialized, function<T>, void (*)(T));

// This function handles all data structures not already handled
template<class T, typename std::enable_if<Is_A_Data_Structure<T>::value, int>::type = 0,
                  typename std::enable_if<!isFunctionExist_Specialized<T>::value, int>::type = 0>
void function(T t) {
    std::cout << "data structure" << std::endl;
    return;
}
// Care, isFunctionExist_Specialized<T>:: value is computed only once,
// so you have to use another
// `HAS_TEMPLATED_FUNC(isFunctionExist, function<T>, void (*)(T));`
// to take into account these new functions.

如何使用SFINAE选择最接近的匹配类型特征？

3 个答案: