我有一个带有一些模板专门化的可变模板类。对于某些特殊化，我希望有条件地启用它们(使用std::enable_if，所以我添加了一个额外的未使用的模板参数(。
下面是一个代码示例：
#include <tuple>
#include <iostream>
template<int v,typename, typename...>
struct A;
template<int v, typename...Ts, typename...Us>
struct A<v,void,std::tuple<Ts...>, Us...> {
void print() {
std::cout << "A: Tuple selected" << std::endl;
}
};
template<int v, typename T, typename...Us>
struct A<v,void,T, Us...> {
void print() {
std::cout << "A: one element selected" << std::endl;
}
};
template<int v,typename, typename...>
struct B;
template<int v, typename...Ts, typename...Us>
struct B<v,std::enable_if_t<v!=11>,std::tuple<Ts...>, Us...> {
void print() {
std::cout << "B: Tuple selected" << std::endl;
}
};
template<int v, typename T, typename...Us>
struct B<v,std::enable_if_t<v==12>,T, Us...> {
void print() {
std::cout << "B: one element selected" << std::endl;
}
};
template<int v, typename...>
struct C;
template<int v, typename...Ts, typename...Us>
struct C<v,std::enable_if_t<v!=11,std::tuple<Ts...>>, Us...> {
void print() {
std::cout << "C: Tuple selected" << std::endl;
}
};
template<int v, typename T, typename...Us>
struct C<v, T, Us...> {
void print() {
std::cout << "C: one element selected" << std::endl;
}
};
int main()
{
struct A<12,void,std::tuple<int>> a;
a.print();
struct B<12,void,std::tuple<int>> b;
b.print();
struct C<12,std::tuple<int>> c;
c.print();
return 0;
}
a的实例化有效(并且确实选择了tuple专用化(。
由于模板安装不明确，b的实例化失败。使用gcc:
$ g++ -std=gnu++17 -Wall -Wextra toto5.cpp
toto5.cpp: In function ‘int main()’:
toto5.cpp:61:38: error: ambiguous template instantiation for ‘struct B<12, void, std::tuple<int> >’
61 | struct B<12,void,std::tuple<int>> b;
| ^
toto5.cpp:25:8: note: candidates are: ‘template<int v, class ... Ts, class ... Us> struct B<v, typename std::enable_if<(v != 11), void>::type, std::tuple<_UTypes ...>, Us ...> [with int v = 12; Ts = {int}; Us = {}]’
25 | struct B<v,std::enable_if_t<v!=11>,std::tuple<Ts...>, Us...> {
| ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
toto5.cpp:32:8: note: ‘template<int v, class T, class ... Us> struct B<v, typename std::enable_if<(v == 12), void>::type, T, Us ...> [with int v = 12; T = std::tuple<int>; Us = {}]’
32 | struct B<v,std::enable_if_t<v==12>,T, Us...> {
| ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
toto5.cpp:61:38: error: aggregate ‘B<12, void, std::tuple<int> > b’ has incomplete type and cannot be defined
61 | struct B<12,void,std::tuple<int>> b;
| ^
对于C，读取模板本身时发生错误(无需实例化(：
$ g++ -std=gnu++17 -Wall -Wextra toto5.cpp
toto5.cpp:42:8: error: template parameters not deducible in partial specialization:
42 | struct C<v,std::enable_if_t<v!=11,std::tuple<Ts...>>, Us...> {
| ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
toto5.cpp:42:8: note: ‘Ts’
能做些什么？
注意：我知道std::enable_if中的条件没有用处，但这只是这里的一个例子。
一种方法是使用B解决方案，但更改std::enable_if条件以禁用第二个模板专用化，但如何说"；不是任何类型的元组"；？std::is_same在这里似乎没有用处。
clang++在所有这些情况下都会给我同样的错误。
对于特殊化B，您需要确保std::enable_if的条件是正交的。在您的版本中，如果您提供v=12，则条件v!=11和v==12都会产生true，这意味着两个版本都已启用。这就是为什么您会得到不明确的实例化错误的原因。以下内容编译良好(https://godbolt.org/z/csaTWMd9v)：
#include <tuple>
#include <iostream>
template<int v,typename, typename...>
struct B;
template<int v, typename...Ts, typename...Us>
struct B<v,std::enable_if_t<v!=11>,std::tuple<Ts...>, Us...> {
void print() {
std::cout << "B: Tuple selected" << std::endl;
}
};
template<int v, typename T, typename...Us>
struct B<v,std::enable_if_t<v==11>,T, Us...> {
void print() {
std::cout << "B: one element selected" << std::endl;
}
};
int main()
{
struct B<12,void,std::tuple<int>> b;
b.print();
}
更新：正如评论中所问的，检查某个模板参数是否是任何类型的元组的方法可以如下所示(例如，也可以比较这篇文章(。这更接近你想要实现的目标吗？(https://godbolt.org/z/dq85fM7KK)
#include <tuple>
#include <iostream>
template <typename>
struct is_tuple : std::false_type
{ };
template <typename... T>
struct is_tuple<std::tuple<T...>> : std::true_type
{ };
template<int v,typename, typename...>
struct B;
template<int v, typename...Ts, typename...Us>
struct B<v, std::enable_if_t<v!=11>, std::tuple<Ts...>, Us...> {
void print() {
std::cout << "B: Tuple selected" << std::endl;
}
};
template<int v, typename T, typename...Us>
struct B<v,std::enable_if_t<v==12 && !is_tuple<T>::value>, T, Us...> {
void print() {
std::cout << "B: one element selected" << std::endl;
}
};
int main()
{
B<12, void, std::tuple<int>>{}.print(); // Tuple selected
B<12, void, int>{}.print(); // one element selected
}