I'm new in the C++ world. Sorry for my nooby question.
I have a class
template <typename T>
class Foo
{
T t_;
void say_hello()
{ std::cout << "Ciao";}
// work with T ...
};
I want to specialize this template class for 3 types.
If type is (A or B or C), Then use this class
template<>
class Foo<A or B or C>
{
void say_hello()
{ std::cout << "Hello";}
};
What's the best way to do this? Thank you for your help.
A possible solution uses SFINAE
template <typename T, typename = void>
class Foo
{
T t_;
void say_hello()
{ std::cout << "Ciao";}
// work with T ...
};
template <typename T>
class Foo<T, std::enable_if_t<std::is_same_v<T, A>,
|| std::is_same_v<T, B>,
|| std::is_same_v<T, C>>
{
void say_hello()
{ std::cout << "Hello";}
};
If you don't use T inside the Foo specialization (as in your example) you can also use a sort of self-inheritance
template <typename T>
class Foo
{
T t_;
void say_hello()
{ std::cout << "Ciao";}
// work with T ...
};
template <>
class Foo<A>
{
void say_hello()
{ std::cout << "Hello";}
};
template <>
class Foo<B> : public Foo<A>
{ };
template <>
class Foo<C> : public Foo<A>
{ };
Off Topic: if you want to use say_hello() outside the class, is better if you make it public (or if you declare Foo as a struct).
There are several possibilities, for example:
Specialization of the method only:
template<>
void Foo<A>::say_hello() { std::cout << "Hello"; }
template<>
void Foo<B>::say_hello() { std::cout << "Hello"; }
template<>
void Foo<C>::say_hello() { std::cout << "Hello"; }
or, in C++17, you might do:
template <typename T>
class Foo
{
T t_;
void say_hello()
{
if constexpr(std::is_same_v<T, A> || std::is_same_v<T, B> || std::is_same_v<T, C>) {
std::cout << "Hello";
} else {
std::cout << "Ciao";
}
}
// work with T ...
};
Whereas regular if works in that example, it would fail if you call code specific to A, B, C.
if constexpr won't have that issue.
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