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當提供operator<<
模板類的專業化,鐺喜歡的朋友內嵌形式:提供適當的運算符<<與可變參數模板專業化
#include <iostream>
template <typename A, typename... Bs>
struct Hello;
template <typename A>
struct Hello<A> {
template <typename A2>
friend std::ostream & operator<<(std::ostream & s, const Hello<A2> & h) {
return s << "specialized\n";
}
};
template <typename A, typename... Bs>
struct Hello {
template <typename A2, typename... B2s>
friend std::ostream & operator<<(std::ostream & s, const Hello<A2,B2s...> & h) {
return s << "generic\n";
}
};
int main()
{
std::cout << Hello<int>()
<< Hello<float>()
<< Hello<int,int>()
<< Hello<int,float>();
}
http://coliru.stacked-crooked.com/a/47743db96c0f3a02
但是GCC失敗的是,它更喜歡非在線版本:
#include <iostream>
template <typename A, typename... Bs>
struct Hello;
template <typename A>
struct Hello<A> {
template <typename A2>
friend std::ostream & operator<<(std::ostream & s, const Hello<A2> & h);
};
template <typename A, typename... Bs>
struct Hello {
template <typename A2, typename... B2s>
friend std::ostream & operator<<(std::ostream & s, const Hello<A2,B2s...> & h);
};
template <typename A2>
std::ostream & operator<<(std::ostream & s, const Hello<A2> & h) {
return s << "specialized\n";
}
template <typename A2, typename... B2s>
std::ostream & operator<<(std::ostream & s, const Hello<A2,B2s...> & h) {
return s << "generic\n";
}
int main()
{
std::cout << Hello<int>()
<< Hello<float>()
<< Hello<int,int>()
<< Hello<int,float>();
}
http://coliru.stacked-crooked.com/a/45328f7bbdb36598
而這又不被叮噹接受。
所以我的問題是:
- 都是形式標準C++?
- 是否有一種表單被編譯器接受?
某種聯繫:http://stackoverflow.com/questions/1297609/overloading-friend-operator-for-template-class – DarioP