[英](Partially) specializing a non-type template parameter of dependent type
Maybe I'm tired, but I'm stuck with this simple partial specialization, which doesn't work because non-type template argument specializes a template parameter with dependent type 'T'
:也许我累了,但我坚持使用这个简单的部分专业化,这不起作用,因为
non-type template argument specializes a template parameter with dependent type 'T'
:
template <typename T, T N> struct X;
template <typename T> struct X <T, 0>;
Replacing 0
by T(0)
, T{0}
or (T)0
doesn't help.用
T(0)
、 T{0}
或(T)0
替换0
没有帮助。 So is this specialization even possible?那么这种专业化甚至可能吗?
See paragraph [temp.class.spec] 14.5.5/8 of the standard:参见标准的 [temp.class.spec] 14.5.5/8 段:
The type of a template parameter corresponding to a specialized non-type argument shall not be dependent on a parameter of the specialization.
对应于专门化的非类型参数的模板参数的类型不应依赖于专门化的参数。 [ Example:
[示例:
template <class T, T t> struct C {}; template <class T> struct C<T, 1>; // error template< int X, int (*array_ptr)[X] > class A {}; int array[5]; template< int X > class A<X,&array> { }; // error
—end example ]
—结束示例]
The answer to your edit: the easiest workaround is to replace a non-type template parameter with a type one:编辑的答案:最简单的解决方法是用类型一替换非类型模板参数:
#include <type_traits>
template <typename T, typename U>
struct X_;
template <typename T, T N>
struct X_<T, std::integral_constant<T, N>> {};
template <typename T>
struct X_<T, std::integral_constant<T, 0>> {};
template <typename T, T N>
struct X : X_<T, std::integral_constant<T, N>> {};
Solution using Yakk's solution:使用 Yakk 的解决方案的解决方案:
#include <iostream>
#include <type_traits>
template <typename T, T N, typename = void >
struct X {
static const bool isZero = false;
};
template <typename T, T N>
struct X < T, N, typename std::enable_if<N == 0>::type > {
static const bool isZero = true;
};
int main(int argc, char* argv[]) {
std::cout << X <int, 0>::isZero << std::endl;
std::cout << X <int, 1>::isZero << std::endl;
return 0;
}
您可以在template
参数列表的末尾添加一个typename=void
参数,然后在专业化中使用std::enable_if_t<
condition >
疯狂。
You need to pass an integral value in a template, Both, your first and second template, will not work if the type T is not an integral type.您需要在模板中传递一个整数值,如果类型 T 不是整数类型,则您的第一个和第二个模板都将不起作用。
You can pass Traits as a typed template parameter to specify the value N:您可以将 Traits 作为类型化模板参数传递以指定值 N:
#include <iostream>
// error: ‘double’ is not a valid type for a template non-type parameter
template <typename T, T N> struct X0;
// error: ‘double’ is not a valid type for a template non-type parameter
template <typename T, T N, int = 0> struct X1;
template <typename T, T N>
struct IntegralTraits {
static constexpr T Value() { return N; }
};
template <typename T, typename Traits = void>
struct X2 {
static constexpr T Value() { return Traits::Value(); }
};
template <typename T>
struct X2<T, void> {
static constexpr T Value() { return T(); }
};
int main() {
// error: ‘double’ is not a valid type for a template non-type parameter
// X0<double, 0>();
// error: ‘double’ is not a valid type for a template non-type parameter
// X1<double, 0>();
X2<int> a;
X2<double, IntegralTraits<int, 1>> b;
std::cout.precision(2);
std::cout << std::fixed << a.Value() << ", "<< b.Value() << '\n';
return 0;
}
If you limit yourself to integral types pick a large one:如果您限制自己使用整数类型,请选择一个大的类型:
template <typename T, std::size_t N = 0> struct X {};
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