[英]Find template type of a template type c++
I would like to have a function that can take many different things (for simplicity) like so: 我希望有一个可以接受许多不同功能(为简单起见)的函数,如下所示:
template <typename T>
typename type_to_return<T>::type // <-- Use type_to_return to get result type
foo(T t)
{
return typename type_to_return<T>::type(T); // <-- Build a thing!
}
I would then specialize the type_to_return
class for the types I have created. 然后,我将专门针对我创建的类型使用type_to_return
类。 This would make the entry be one function and I could then just define new type_to_return
s and constructors. 这将使该条目成为一个函数,然后我就可以定义新的type_to_return
和构造函数。
I want type_to_return<T>::type
to be just T
if T
is not some class template. 如果T
不是某些类模板,我希望type_to_return<T>::type
只是T
Otherwise I want it to be that class's first template parameter. 否则,我希望它成为该类的第一个模板参数。 So for int
, I get back int
, and for MultOp<float,int,double>
, I want to get back float
. 因此对于int
,我返回int
,对于MultOp<float,int,double>
,我想返回float
。
How do I do that? 我怎么做? I think I need to do something like: 我想我需要做些类似的事情:
// Base version
template <typename T>
struct type_to_return
{
typedef T type;
};
// Specialized type
template <template <typename> class T>
struct type_to_return<T <any_type_somehow> >
{
typedef template boost::magic_type_unwrapper<T>::param<1>::type type;
};
You may implement a type_unwrapper
as follow: 您可以如下实现type_unwrapper
:
template <typename T>
struct type_unwrapper;
template <template <typename...> class C, typename... Ts>
struct type_unwrapper<C<Ts...>>
{
static constexpr std::size_t type_count = sizeof...(Ts);
template <std::size_t N>
using param_t = typename std::tuple_element<N, std::tuple<Ts...>>::type;
};
which works as long there is no template value as in std::array<T, N>
. 只要没有std::array<T, N>
模板值, std::array<T, N>
。
Note also that stl container declare some typedef
to retrieve there template arguments as std::vector<T, Alloc>::value_type
which is T
. 还要注意,stl容器声明一些typedef
来检索模板参数,如std::vector<T, Alloc>::value_type
,即T
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