如果需要参数包,则使用SFINAE进行特殊化 [英] Specializing class with SFINAE if a parameter pack is needed
问题描述
当我对这个问题有了一个完美的答案时:
使用SFINAE进行专业培训的班级
As I got a perfect answer for the question: Specializing class with SFINAE
为完整起见,我在此处再次插入正确的解决方案作为示例:
For completeness I insert the correct solution as example here again:
class AA { public: using TRAIT = int; };
class BB { public: using TRAIT = float; };
template < typename T, typename UNUSED = void> class X;
template < typename T >
class X<T, typename std::enable_if< std::is_same< int, typename T::TRAIT>::value, void >::type>
{
public:
X() { std::cout << "First" << std::endl; }
};
template < typename T >
class X<T, typename std::enable_if< !std::is_same< int, typename T::TRAIT>::value, void >::type>
{
public:
X() { std::cout << "Second" << std::endl; }
};
int main()
{
X<AA> a;
X<BB> b;
}
但是如果我必须使用参数包进行进一步使用,我看不到有机会写下这样的东西:
But if I have to use a parameter pack for further use, I see no chance to write the things down like:
template < typename T, typename ...S, typename UNUSED = void> class X;
错误:参数包 S必须位于模板参数列表
error: parameter pack 'S' must be at the end of the template parameter list
以不同的顺序定义
template < typename T, typename UNUSED = void, typename ...S> class X;
如果使用第一种其他类型,则会出现问题。
ends up in problems if the first additional type is in use.
好的,我所描述的是我实际上找不到的技术解决方案。也许有另外一种。我的基本问题是什么:该类需要2个不同的构造函数,这些构造函数调用不同的基类构造函数。但是,因为两个构造函数都具有相同的参数集,所以我没有机会专门化构造函数本身。
OK, what I describe is a technical solution which I can't find actually. Maybe there is a different one. What is my underlying problem: I need 2 different constructors for the class which call different base class constructors. But because both constructors have the same set of parameters I see no chance to specialize the constructors itself.
如果专门化构造函数可以工作,则可能是这样的:
If specialize constructors can work, it can be something like that:
template < typename T>
class Y
{
public:
template <typename U = T, typename V= typename std::enable_if< std::is_same< int, typename U::TRAIT>::value, int >::type>
Y( const V* =nullptr) { std::cout << "First" << std::endl; }
template <typename U = T, typename V= typename std::enable_if< !std::is_same< int, typename U::TRAIT>::value, float >::type>
Y( const V* =nullptr) { std::cout << "Second" << std::endl; }
};
错误:'模板Y :: Y(const V *)'无法重载
error: 'template template Y::Y(const V*)' cannot be overloaded
但是如上所述。 ..我不知道是否可以做到这一点。
But as already mentioned... I have no idea if that can be done.
为了显示潜在的问题,我将给出以下示例,该示例显示了依赖于基类构造函数的不同用法
To show the underlying problem, I would give the following example which shows the different use of base class constructors dependent on a trait which is defined in the base class.
template <typename T, typename ... S>: public T
class Z
{
public:
// should work if T defines a trait
Z( typename T::SomeType t): T( t ) {}
// should be used if T defines another trait
Z( typename T::SomeType t): T( ) {}
};
推荐答案
代替
template < typename T, typename ...S, typename UNUSED = void> class X;
您可以添加一层:
template <typename T, typename Dummy = void, typename ... Ts> class X_impl {};
然后
template <typename T, typename ...Ts>
using X = X_impl<T, void, Ts...>;
对于SFINAE,因为默认模板参数不是签名的一部分,
For SFINAE, as default template parameter is not part of signature,
template <typename U = T,
typename V = std::enable_if_t<std::is_same<int, typename U::TRAIT>::value, int>>
Y(const V* = nullptr) { std::cout << "First" << std::endl; }
template <typename U = T,
typename V = std::enable_if_t<!std::is_same<int,
typename U::TRAIT>::value, float>>
Y(const V* = nullptr) { std::cout << "Second" << std::endl; }
应重写,例如:
template <typename U = T,
std::enable_if_t<std::is_same<int, typename U::TRAIT>::value>* = nullptr>
Y() { std::cout << "First" << std::endl; }
template <typename U = T,
std::enable_if_t<!std::is_same<int, typename U::TRAIT>::value>* = nullptr>
Y() { std::cout << "Second" << std::endl; }
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